M,    ^-^.^u  ."^_V,  -"--^ 


SECOND  COURSE  IN  ALGEBRA 


rif.ii^L 


BY 


HERBEET  E.  HAWKES,  Ph.D. 

PROFESSOR  OF  MATHEMATICS  IN  COLUMBIA  UNIVERSITY 


WILLIAM  A.  LUBY,  A.B. 


EEANK  C.  TOUTON,  Ph.B. 

INSTRUCTORS  IN  MATHEMATICS  IN  CENTRAL  HIGH  SCHOOL 
KANSAS  CITY,  MISSOURI 


■N. 


u 


OF  THE 


OF 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


?- 


'U 


Copyright,  1911,  by 

Hbbbebt  E.  Hawkes,  William  A.  Luby,  and 
Frank  C.  Touton 


ALL  BIGHTS  RESERVED 
911.5 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

This  book  is  designed  to  follow  the  authors'  "  First  Course 
in  Algebra  "  or  any  other  text  of  similar  scope  and  treatment. 
Experience  shows  that  when  a  student  returns  to  the  study 
of  algebra,  after  even  a  summer's  vacation,  a  review  is  very 
necessary ;  and  that  it  is  absolutely  indispensable  if  he  comes 
back  after  a  year  spent  on  geometry.  The  review  presented 
in  the  early  chapters  is  brief,  yet  sufficiently  thorough.  Each 
review  topic  has  been  given  a  broader  and  more  advanced  treat- 
ment than  is  permissible  in  a  first  course.  New  material  is  used 
throughout  and  many  new  applications  are  given  in  order 
to  make  the  entire  review  appeal  to  the  student  as  fresh  and 
inviting. 

In  the  chapters  which  deal  with  the  subjects  not  given  in  the 
"  First  Course,"  the  aim  has  been  to  select  those  topics  con- 
sidered necessary  for  the  best  secondary  schools  and  to  treat 
each  in  a  clear,  practical,  and  attractive  manner.  It  has  been 
the  purpose  also  to  prepare  a  text  that  will  lead  the  student  to 
think  clearly,  as  well  as  to  acquire  the  necessary  facility  on  the 
technical  side  of  algebra.  Lastly,  it  has  been  the  desire  to  re- 
duce the  work  of  explanation  and  illustration  on  the  teacher's 
part  to  a  minimum.  To  accomplish  these  things  every  legiti- 
mate resource  has  been  employed.  The  material  has  been  care- 
fully selected  and  graded,  the  explanations  are  unusually  full, 
and  the  illustrative  examples  are  especially  numerous.  When- 
ever graphs  appeared  to  clarify  a  subject,  they  have  been  used ; 
and  if  at  any  point  an  explanatory  note  or  a  bit  of  mathemati- 
cal history  seemed  pertinent,  it  has  been  given.  Along  with 
the  endeavor  to  accomplish  these  various  ends  a  continuous 


vi  SECOND  COURSE  IN  ALGEBRA 

effort  has  been  made  to  produce  a  text  that  is  modern,  lucid, 
mathematically  correct,  and  interesting. 

We  are  under  especial  obligation  for  suggestions  and  criti- 
cisms, not  only  to  those  mentioned  in  the  preface  to  our  "  First 
Course,"  but  also  to  Mr.  J.  M.  McPherron  of  Los  Angeles, 
California ;  Mr.  J.  A.  Foberg  of  Chicago,  Illinois  ;  and  Mr.  A.  E. 
Booth  of  New  Haven,  Connecticut,  who  have  read  the  manu- 
script critically.  For  a  careful  reading  of  the  entire  proof  and 
for  many  helpful  comments  we  are  indebted  to  Mr.  J.  A.  Avery 
of  Somerville,  Massachusetts,  and  Professor  I.  M.  De  Long  of 
Boulder,  Colorado. 


CONTENTS 

CHAPTER  PAGE 

I.   Fundamental  Operations  (Sects.  1-10)    ....  1 

II.   Factoring  (Sects.  11-27) 16 

III.  Fractions  (Sects.  28-34) 34 

IV.  Linear  Equations  in  One  Unknown  (Sects.  3.5-37)  42 
Y.   Linear  Systems  (Sects.  38-47) 54 

VI.   Roots,  Radicals,  and  Exponents  (Sects.  48-62)    .  82 
VII.   Graphical  Solution  of  Equations  in  One  Un- 
known (Sects.  63-69) 110 

VIII.   Quadratic  Equations  (Sects.  70-71) 118 

IX.   Irrational  Equations  (Sect.  72) 128 

.     X.   Graphs  of  Quadratic  Equations  in  Two  Vari- 
ables (Sects.  73-74) .135 

XL   Systems  Solvable  by  Quadratics  (Sects.  75-81)  143 

XIL   Progressions  (Sects.  82-91) 159 

XIII.  Limits  and  Infinity  (Sects.  92-97) 176 

XIV.  Logarithms  (Sects.  98-112) 183 

XV.   Ratio,  Proportion,  AND  Variation  (Sects.  113-118)  207 

XVI.   Imaginaries  (Sects.  119-126) 222 

XVII.   Theory  of  Quadratic  Equations  (Sects.  127-131)  234 

XVIII.   The  Binomial  Theorem  (Sects.  132-137)  ....  242 

XIX.   Supplementary  Topics  (Sects.  138-142)    ....  249 

INDEX • 263 


vu 


ILLUSTRATIONS 


PAGR 

GOTTFRIED  WILHELM  LEIBNITZ 64 

FELIX  KLEIN 116 

JOHN  NAPIER 190 


viii 


SECOND  COURSE  m  ALGEBRA 


CHAPTER  I 

FUNDAMENTAL  OPERATIONS 

1.  Order  of  fundamental  Operations.  The  numerical  value  of  an 
arithmetical  or  an  algebraic  expression  involving  signs  of  addi- 
tion, subtraction,  multiplication,  and  division  depends  on  the 
order  in  which  the  indicated  operations  are  performed.  It  is 
understood  that : 

In  a  series  of  operations  involving  addition,  subtraction,  mul- 
tiplication, and  division,  first  the  multiplications  and  divisions 
shall  be  perform,ed  in  the  order  in  which  they  occur.  Then  the 
additions  and  subtractions  shall  be  performed  in  the  order  in 
which  they  occur  or  in  any  other  order. 

Within  any  parenthesis  the  preceding  rule  applies. 

EXERCISES 

Simplify : 

1.  3  _  5  _}_  6  -  8.         3.  24  --  8  •  4  -  4  +  6. 

2.  6-2  +  1^4.         4.  (7-6)(18-2.4)-4-(20--4). 

5.  42-2(18-2.3)-i-4  +  3.5J 

6.  16  +  4- 8-10  + 51 -16- 4- 6. 3- 0-2  + 18.  8 

-48-2  18-^-12. 

7.  (16  -  32  X  48  -  8  -  4  -  8  +  3)  X  [12  -  4  -  3  -  1] 

+  (42  -  6  •  7  -  42  -  6)  .  6. 

8.  Dors  a*  =  4 a  when  a  =  3  ?  when  a  =  2?  when  o^  =  0 ? 

9.  What  name  is  given  to  each  4  in  a*  =  4  a  ?  Define  each. 
10.  Define  power*  Distinguish  between  exponent  and  power. 


•     2  SE(;OTO  COURSE  IN  ALGEBRA 

,Fi  ad  ^WliufiiericajV.value  of: 

11.  x^  —  5x  -\-  6  when  ic  =  5. 

12.  a;»  -  3a;2  +  Sa-  -  1  when  x  =  S. 

13.  x^  —  S  x^y  -\- S  xi/ —  i/  when  x  =  4t  and  ?/  =  2. 

14.  .,  :    ^ -^-^  when  aj  =  3  and  ?/ =  2. 

x^  —  xy  +  f        x-\-y  -^ 

15.  What  is  the  absolute  value  of  a  number  ?    Illustrate. 

2.  Addition.  In  algebra,  addition  involves  the  uniting  of 
similar  terms  (see  definitions  below)  which  have  the  same  or 
opposite  signs  into  one  term.    For  this  we  have  the  rule : 

I.  To  add  two  or  inore  positive  numbers,  find  the  arithmetic 
cal  swm  of  their  absolute  values  and  prefix  to  this  sum  the 
plus  sign. 

II.  To  add  two  or  more  negative  numbers,  find  the  arithmeti- 
cal sum,  of  their  absolute  values  and  prefix  to  this  sum  the 
minus  sign. 

III.  To  add  a  positive  and  a  negative  number,  find  the  differ- 
ence of  their  absolute  values  and  prefix  to  this  difference  the  sign 
of  the  number  which  has  the  greater  absolute  value. 

Obviously  24-4  +  7  =  2  +  7  +  4  =  7  +  2  +  4,  etc.  Even  if 
we  have  a  series  of  positive  and  negative  numbers,  the  order  in 
which  they  occur  does  not  affect  the  final  result.  This  princi- 
ple of  addition  is  called  the  Commutative  Law  for  Addition. 

Similar  terms  are  (a)  integers  and  rational,  numerical  frac- 
tions; (b)  like  indicated  roots,  as  V2  and  3  V2,  or  V3  and  2  VS; 
(c)  terms  having  like  literal  parts,  as  4  a  and  ^a,  or  Q>xy'^  and 

V2V- 

Dissimilar  terms  are  unlike  indicated  roots  or  terms  having 
unlike  literal  parts. 

For  the  addition  of  polynomials  we  have  the 

Rule.    Write  similar  terms  in  the  same  column. 
Find  the  algebraic  sum  of  the  terms  in  each  colum^n  and  write 
the  results  in  succession  with  their  proper  signs. 


FUNDAMENTAL  OPERATIONS  B 

3.  Subtraction.  For  the  subtraction  of  polynomials  we  have 
the  following 

Rule.  W^ite  the  subtrahend  under  the  minuend  so  that  simi- 
lar terms  are  in  the  same  columyn. 

Change  mentally  the  sign  of  each  term  of  the  subtrahend. 
Then  find  the  algebraic  sum  of  the  terms  in  each  column,  and 
write  the  results  in  succession  with  their  proper  signs. 

EXERCISES  *    /  -^ 

Add: 

1.  16,  -  3,  +  2,  -  8,  -  7,  and^r 

2.  4  «,  —  6  «,  —  10  «^,  4-  2  a,  and  18  a. 

3.  4ic-3  2/  +  7,  8a;-10?/-ll,  andl02/-30  +  7cc. 

4.  7  ic  —  4  2/  —  ,-?,  3  £c  +  .-^  —  8  ?/,  and  ISy  —  17  x  —  14:Z. 

5.  4.a^-3a^c-4.ac^,  S  a^c  -  Sac^  -  Sa%  and  Sa^-6a^c. 

6.  It  X  =  1,  y  =  2,  and  z  =  S,  find  the  numerical  value  of 
each  of  the  three  expressions  and  of  the  result  obtained  in 
Exercise  4.  Compare  the  sum  of  the  three  numerical  values 
with  the  numerical  value  of  the  result. 

7.  State  a  rule  for  checking  work  in  addition  of  algebraic 
expressions. 

Write  with  polynomial  coefficients  : 

8.  ay-{-by-\-  cy.  12.  S(a  +  b)— c(a -{-  b). 

9.  ^ ax  -  4.bx  +  Q> X.      13.  Qa(x  —  2c)- ^(x  —  2c). 

10.  4:x-abx-x.  U.  4.b(Sx-2)-Sc(Sx-2). 

11.  7x  —  3ax-4.a^x.     15.  4.m(5 a-S c)- 6n(—3c-\-5  a). 

Subtract  the  first  expression  from  the  second  in : 

16.  Aa,6a.  18.  4. x -\- S,  S x -\- 6. 

17.  8  a%  5  a^  19.  7  x^  -  10,  5x^  +  20. 

20.  x  —  3if  +  z  —  4:ac  +  7  ax,  4:X  —  y^  -{-  S  —  5  ax  -\- 9 ac. 

21.  a^  —  c  +  Sx  —  ahti  —  8  ac,  4  a^ -\-  m  —  8 a?  — 10  ac  +  4  a^m. 


4         SECOND  COURSE  IN  ALGEBRA 

Find  the  expression  which  added  to  the  first  will  give  the 
second  in : 

22.  x^-  bx  +  6,  ^x""  -bx-\-2. 

23.  4ic2_3^^^2^3^2_,.7^_10rc24.8. 

Find  the  expression  which  subtracted  from  the  first  will 
give  the  second  in : 

24.  4a^-2„a&  +  6^7a2-10aZ»  +  6Z»l 
25.' c^  -iOcx  +  %x^,^x^-10cx  +  4.-\-  c\ 

26.  State  a  rule  for  checking  work  in  subtraction  suggested 
by  the  directions  preceding  Exercises  22  and  24. 

27.  From  the  sum  of  ax  —  ac  —  ^c^  and  4  c^  —  3  ac  take  the 
sum  of  4  c^  —  8  aa;  +  a^  and  4  cm;  -f-  3  ax  —  5  c^ 

Reinove  parentheses  and  combine  like  terms : 

28.  4oc-3-(a-2ic)  +  (3a;-a). 

29.  6a;+(3c-8a;4-2)-(c-ic-2). 

30.  6x-[-(a-c)4-(3c-4a)]. 

31.  7c -[(3c -4)- 6 -(4a;- 3a -c)]. 

32.  4a;-2(a;-3)-3[a;-3(4-2a;)4-8]. 

33.  6a;  -  4(3  -  5a;)-  4[2(a;  -  4)+  3(2a;  -  l)-(a;  -  7)]. 

34.  3a;  -  2[1  -  3(2 a;  -  3  -  a)  -  5  {a  -  (3  a;  -  2  a)  -  4}]. 

35.  State  the  rule  for  the  removal  of  a  parenthesis 

(a)  when  it  is  preceded  by  the  sign  plus ; 

(b)  when  it  is  preceded  by  the  sign  minus. 

Inclose  in  a  parenthesis  preceded  by  the  sign  plus  those 
terms  which  contain  x  and  y,  and  inclose  all  other  terms  in  a 
parenthesis  preceded  by  the  sign  minus. 

36.  x"  +  2xy  Jff-a\  37.  x"  +  14a&  -  49^2  _  h\ 

38.  ?/^4-6a;?/  +  9a;2-m'^-10m-25. 

39.  a;*  +  10a;y  -  c«  +  12c*cZ  -  36c^  +  25/. 

40.  State  the  rules  for  inclosing  terms  in  a  parenthesis  pre- 
ceded by  (a)  the  sign  plus ;  (h)  the  sign  minus. 


FUNDAMENTAL  OPERATIONS  5 

4.  Multiplication.  In  multiplying  one  term  by  another  the 
sign  of  the  product,  the  coefficient  of  the  product,  and  the  expo- 
nent of  any  letter  in  the  product  are  obtained  as  follows  : 

I.  The  sign  of  the  product  is  plus  if  the  multiplier  and  the 
multiplicand  have  like  signs,  and  minus  if  they  have  unlike  signs. 

II.  The  coefficient  of  the  product  is  the  product  of  the  coeffi- 
cients of  the  factors. 

III.  The  exponent  of  each  letter  in  the  product  is  detei^mined 
Uj  the  general  law  ^^^^^^a^h^ 

For  the  multiplication  of  polynomials  we  have  the 
E/ULE.    Multiply  the  multiplicand  by  each  term,  of  the  multi- 
plier in  turn,  and  add  the  partial  products. 

An  extension  of  the  law  for  exponents  in  multiplication  is 
the  Law  of  Involution  :         .    .  j,  _  ^^j, 

This  last  law  implies  the  more  general  forms : 

and  .{{x''fy=x''^', 

5.  Division.  In  dividing  one  term  by  another  the  sign  of  the 
quotient,  the  coefficient  of  the  quotient,  and  the  exponent  of 
any  letter  in  the  quotient  are  obtained  as  follows : 

I.  The  sign  of  the  quotient  is  plus  when  the  dividend  and  the 
divisor  have  like  signs,  and  minus  when  they  have  unlike  signs. 

II.  The  coefficient  of  the  quotient  is  the  quotient  of  the  coeffi^ 
cient  of  the  dividend  by  that  of  the  divisor. 

III.  The  exponent  of  each  letter  in  the  quotient  is  determined 
by  the  law  _        ^        „_j. 

The  method  of  dividing  one  j)olynomial  by  another  is  stated 
in  the 

Rule.  Arrange  the  dividend  and  the  divisor  according  to  the 
descending  powers  of  some  common  letter,  called  the  letter  of 
airrangement. 


G  secum:)  course  is  aluebra 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor  and  write  the  result  for  the  first  term  of  the  quotient. 

Multiply  the  entire  divisor  by  the  first  term  of  the  quotient, 
write  the  result  under  the  dividend,  and  subtract,  being  careful 
to  write  the  terms  of  the  remainder  in  the  same  order  as  those 
of  the  divisor. 

Divide  the  first  term  of  the  remainder  by  the  first  term,  of  the 
divisor  for  the  second  teryn^ofihe  quotient,  and  proceed  as  before 
until  there  is  no  remainder,  or  until  the  remainder  is  of  lower 
degree  (§8)  in  the  letter  of  arrangement  than  the  divisor. 

6.  Meaning  of  a  zero  exponent.  The  laws  for  exponents  stated 
in  the  formulas  of  §  §  4  and  5  are  assumed  to  hold  for  all  values 
of  a  and  b. 

Then  ic"  ^  x«  =  x"-"  =  x^. 

But  -  =  1. 

Hence  x'*  =  \. 

That  is,  any  number  (except  zero)  whose  exponent  is  zero  is 
equal  to  1. 

7/  Meaning  of  a  negative  exponent.  If,  in  the  formula  of  §  5, 
b  is  greater  than  a,  we  obtain  a  negative  exponent  for  n.  The 
meaning  of  such  an  exponent  is  illustrated  as  follows : 


y 


But  dividing  both  terms  by  cc", 

a;«  ic"   _  1 

Therefore  x~^  =  — 

x^ 

C  *         C  '         '" 

More  generally,  cx~  °  =  —  ,  and =  cx°. 

x^  x-° 

Hereafter  it  will  be  assumed  that  all  the  preceding  exponential 
laws  hold  for  positive,  negative,  zero,  and  fractional  exponents. 


FUNDAMENTAL  OPERATIONS  7 

EXERCISES 
Perforin  the  indicated  operation  : 

1.   (4.x^-Sx)(2x).  2.  (2ic  +  3)(5ic-6). 

3.  Substitute  2  for  x  in  each  of  the  factors  of  Exercise  2 , 
and  in  the  product.  Compare  the  numerical  value  of  the  product 
with  the  product  of  the  numerical  values  of  the  factors.  Then 
state  a  method  of  checking  numerically  work  in  multiplication. 

4.  (3ic2-l)2.  8.   (e'-^e-^y. 

5.  (7ic2«-8ic«  +  3)2.  9.  (e^+.2e-*)l 

6.  (xi-\-x^y.  Q'         10.   (e^-e-'^f. 

7.  (x^  -xf.  11.   (e2--3e-^)^ 

12.  (x^  +  X?  +  1)  (x^  -x^  +  1). 

-(¥-?— ){¥-f-i)- 

14.  (4x3«  -6x'-\-  S)(Jx^'  -  x^'  +  4). 

0  15.  (x^  -  2x2/  -{-  y')(x^  +  2xf  +  i/). 

16.  (x-''-3x-2x-y. 

17.  (x--2  -\-2x^  -Sx^y. 

,„    /2  a''      a       2\/2a^      «2       2a\ 

19.   (5a;2«-3^-2«-6x-«  +  3£i'«)2. 

20.  x^  -  ic  -  90  =  ?  if  X  =  -  9. 

21.  x''  -  4  V  +  4^^  =  ?  if  X  =  3  and  2/  =  2. 

22.  £c^  -  3icV  +  3  V  -  /  =  ?  if  ^  =  2  and  ?/  =  -  3. 

23.  x^  +.Sxhj  -\-  3xi/  +  y^  =  ?  if  ic  =- 4  and  3/ =  -  2. 

24.  (gi  _  e-i)2  =  ?  if  e  =  2  ;  if  e  =  -  3. 

25.  e^^  -  2  e°  +  e-2-  =  ?  if  e  =  2  and  ic  =  2. 

26.  (8£c*-6ic2_4cc)--(-2£c). 

27.  (x^-7x-^l2)^(x-S)^ 

.     28.  State  the  Associative  Law  of  Multiplication.    Illustrate. 
29.  State  the  Distributive  Law  of  Multiplication.    Illustrate. 


8  SECOND  COURSE  IN  ALGEBUA 

30.  (a;»  -  64:)-^(x  -  4)(x'  -  4a;  +  16). 

31.  (x'  -  Sx^  -{-  S3  X  -  SO)-i-(x^  +  Sx  -  5). 

32.  State  a  method  of  checking  work  in  division  similar  to 
the  check  of  multiplication. 

Find  the  remainder  in : 

33.  (Sx""  -  x^  -  6)-h(2x  -  S). 

34.  (4a;*  -x^-  S)-i-(2x^  -x-1). 

Divide : 

35.  ic»  +  8/  +  125  -  SOxy  by  a;  +  2^/  +  5. 

36.  x^  -\- y^  -\-  z^  —  S  xyz  by  x  4-^  +  z. 

37.  a^  -  ab^  +  Jb  -  b^  by  J  -  bK 

38.  Sx-^'^  +  x'^-4:X-'^hy2x-^  +  x^-\-3x-\ 

39.  x^  —  y^  by  [(x^  —  y^)  -4-  (x^  +  y^)]. 

40.  9m  +  4m-i-13  by  3m^-5H-27yri. 
•41.  a;2«  -h  4a;-2«  —  29  by  a;"  -  2a;-«  -  5. 

a 

42.  9a;2«  +  25a;-*«  -  19a;-«  by  5a;-2«  +  3a;«  -  7a;"^. 
"■  (I' +  ¥)-(t  +  t)^*'^-^  +  ^'''^^  +  l^*^)• 
„   /9«*  ,   243a'      ,„   ,   69a      443a^      43a'\ 
"i^  +  ^O ^^  +  ^-"30 rj 

8.  Detached  coefficients.  A  ^term  is  rational  if  it  may  be 
obtained  from  unity  and  the  letters  involved  by  means  of  the 
four  fundamental  operations  without  the  extraction  of  any  root. 

A  term  is  integral  if  it  has .  no  literal  denominator  and  the 
exponent  of  each  factor  is  a  positive  number  (or  zero). 

The  degree  of  a  rational  integral  term  is  the  sum  of  the 
exponents  of  the  letters  in  the  term. 


i 


FUNDAMENTAL  OPERATIONS  9 

An  algebraic  expression  is  rational  and  integral  if  its  terms 

e  rational  and  integral. 

An  integral  expression  may  not  be  rational.    Nor  is  every 

rational  expression  integral.     Thus,   —  +  Vx  +-  8   is   integral 

x^      1  , 

but  not  rational,  while  —  -\ h  8  is  rational  but  not  integral. 

A  rational  integral  expression  is  homogeneous  if  its  terms 
are  all  of  the  same  degree. 

In  the  multiplication  or  division  hi  polynomials  which  in- 
volve but  one  letter  or  which  are  homogeneous  in  two  letters 
much  labor  can  be  saved  by  using  the  coefficients  only. 

EXAMPLES 

1.  Multiply  3ic^  -  4a'  +  6  by  2^2  -  5x  +  3. 

Solution  :  Since  x^  is  missing  in  the  first  expression,  its  coefficient 

is  zero.    Inserting  Ox^  and  detaching  coefficients,  the  multiplication 

is  as  follows :  ^   ^     c\      \    ^     a 

o  +    ()  —  4  +    o 

2-5  +  3 


6+    0-8  +  12 
-15  +  0  +  20-30 

+  9  +    0-12  +  18 
6  -^15  +  1  +  32-42+18 

Supplying  the  powers  of  x,  we  obtain  as  the  product  6  x^  —  15  a;* 
2-3  +  32  a,'2  -  42  X  +  18. 

2.  Divide 
6x'  -  llxhj  +-  2xhf  +  21x7/  -  18?/*  by  2x^  -  hxy  -f-  ^tf. 


Solution :  6-11+2  +  27-18 

6  -  15  +  18 


4  -  16  +  27 
4-10  +  12. 

-  6  +  15  - 

-  6  +  15  - 

-18 

-18 

5  +  6 


3+2-3 


Therefore  the  quotient  is  3  .r"-^  +  2  xy  —  3  if-. 

In  both  multiplication  and  division  by  detached  coefficients 
zero  must  be  supplied  for  the  coefficient  of  any  missing  term. 


10  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Use  detached  coefficients  and  perform  the  indicated  operation : 

1.  (x^-Sx  +  16)(2x-3). 

2.  (-X^  -  4:X  -\-  i)(x^  +  4:X  -\-  ^). 

3.  (a'  -  ah  +  b^)(a''  +  ab  +  b''). 

4.  (2a;?  +  5cc^-2)-^(2:c  +  l). 

5.  (x^  +  4:X-16)-h(x-2). 

6.  (3  xf/  -  6  /  -  2  .t7(8  a:^  -  6  /  -  5  icy). 

•     7.   (9x*-4a;  +  13a;2  +  4-6a;^)--r3^2-.r +  2). 

8.  (.•''4-4/)^(a.^-2^y  +  2  2/-0. 

9.  (81  a*  -  171  d'b^  +  25  Z»^)  --  (9  a^  _  5  z,-^  +  9  ab). 

10.  (4  a«  _  2  a^  -  3  a-  2  -  5  «- 1  +  2  tt)  -!-  (2  ^2  _  2  -  a"  i). 

11.  (Sx  -  12a-V  +  6a;V  -  r')^(2^^  -  2/"')- 

12.  Which  expressions  in  the  preceding  exercises  are  (rr)  not 
integral  ?  (Ir)  not  rational  ? 

Note.  It  is  interesting  to  observe  that  our  ordinary  decimal  nota- 
tion really  involves  the  use  of  detached  coefficients.  The  number 
649,  for  instance,  is  an  abbreviated  way  of  writing  6  •  10^  +  4-10  +  0. 
In  fact,  the  various  digits  in  any  number  in  the  decimal  form  are 
the  detached  coefficients  of  some  power  of  the  number  10. 

9.  Synthetic  division.  This  method  of  division  abbreviates 
the  actual  work,  where  the  divisor  is  a  binomial. 

EXAMPLES 

1.  By  long  division  we  have 

Sx^-Sx^  +  9x-    8  (a: -2 


3a;8-6a-2  \^  x^  -  2  x  +  5 


-  2  x2  +  9  a; 

-  2  x2  +  4  X 

5x- 
5x- 

■    8 
10 

2,  Remainder 

The  preceding  division  can  be  shortened  by  omitting  the 
letters  and  arranging  the  work  as  was  done  in  Example  2, 


FUNDAMENTAL  OPERATIONS  11 

page  9.  But  the  usual  process  just  given  can  be  abbreviated 
still  further.  We  may  also  omit  the  first  number  of  each 
partial  product,  since  it  is  merely  a  repetition  of  the  number 
just  above  it.    Thus  we  obtain 

811-2 


10 


2,  Remainder 

Since  the  sign  minus  before  2  in  the  divisor  changes  every 
sign  in  the  partial  products,  if  we  replace  —  2  by  -|-  2,  we  may 
add  the  partial  products  thus  formed  to  the  dividend  instead 
of  subtracting  them.  (This  change  of  sign  is  not  an  actual 
necessity,  but  it  is  a  great  convenience  in  practical  work.)  Then 
bringing  the  figures  into  horizontal  lines  and  using  only  the 
econd  terni  of  the  divisor  with  its  sign  changed,  we  have  a 
further  abbreviation  of  the  process: 

^_8  +  9-    8[2 

+  6-4  +  10 
3  -  2  +  5  +    2 

Here  we  have  all  the  essential  work  for  the  complete  divi- 
J-ion  of  3  x^  —  8  x^  +  9  ic  —  8  by  ic  —  2.  For  the  figures  on  the 
lower  line  3,  —  2,  5  up  to  the  remainder,  2,  are  the  coefficients 
of  the  partial  quotient  3  ic^  —  2  a?  +  5. 

2.  Divide  2a;*  -  l^.x'  -  Q>x  -  54  by  x  +  3. 

Solution  :  2  +  0  -  14  ^    6  -  54 [-3 

-6+18-12  +  54 
2-6+4-18        0 
Therefore  the  quotient  is  2  x^  —  6  x^  +  4  a;  —  18. 
Since   the    remainder,  is    zero,    a?  +  3    is    a   factor    of    the 
dividend.    This  method  of  obtaining  factors  will  be  used  in 
Exercises  5-21,  page  26. 


1-2  SECOND  COURSE  1^  ALGEBRA 

EXERCISES 
Divide  by  synthetic  division : 

1.  ic^  _  ip  _  12  by  a;  +  3. 

2.  ic*- 2a; -4  by  a; -2. 

-  3.  x-«  +  2  a-'^  +  96  by  ic  +  4. 

Using  synthetic  division,  find  the  remainder  in  Exercises  4, 
6,  8,  and  10 : 

5.  Substitute  ti  for  a;  in  a;^  —  5  a;  +  6  and  compare  the  result 
with  the  remainder  obtained  in  Exercise  4. 

6.  (x^  -\-bx  -{-  (i)  H-  (x  —  n). 

7.  Put  n  for  x  in  x^  -\-  bx  ■+■  c  and  again  compare  results. 

8.  (x^  -\-  ax^  -\-hx  -{-  c)  -i-(x  —  n). 

9.  Compare  the  remainder  in  Exercise  8  with  the  result 
obtained  by  substituting  n  for  x  in  x^  +  ax^  +  bx  -\-  c. 

10.  (a;«-2a;2+6)^(a;-4). 

11.  Substitute  4  for  a;  in  a;^  —  2  a;^  +  6  and  compare  the  result 
with  the  remainder  obtained  in  Exercise  10. 

12.  Draw  a  general  conclusion  from  Exercises  4  to  11. 

Note.  The  approximate  solution  of  equations  of  the  third  and 
higher  degrees,  having  numerical  coefficients,  was  a  problem  to  which 
Newton  devoted  considerable  attention.  Little  progress  in  this  line 
was  made  from  his  time  until  1819,  when  William  George  Horner 
(1786-1837),  a  teacher  in  Bath,  England,  published  a  method  of 
solving  equations  by  synthetic  division.  His  procedure  was  in  essence 
very  similar  to  Newton's,  and  its  element  of  originality  lay  in  the 
very  compact  and  elegant  form  in  which  he  arranged  the  numerical 
work.  In  the  ninety  years  which  have  intervened  since  its  publica- 
tion Horner's  method  has  been  improved  but  little,  which  is  rather 
remarkable,  as  Horner  did  not  have  the  advantage  of  a  university 
training  and  was  by  no  means  a  great  mathematician. 

10.  Important  special  products.  Certain  products  are  of  fre- 
quent occurrence.  These  should  be  memorized  so  that  one  can 
write  or  state  the  result  without  the  labor  of  actual  multiplication. 


FUNDAMENTAL  OPERATIONS  13 

I.  Tor  the  square  of  the  sum  of  two  terms  we  have  the 

formula  ,        ,^„        ,  ,       ,„ 

{a  +  &)2  =  a^  +  2  a6  +  6^. 

II.  For  the  square  of  the  difference  of  two  terms  we  have 
the  formula  ^        ,^„        ,      ^    , 

III.  For  the  product  of  the  sum  and  the  difference  of  two 
terms  we  have  the  formula 

IV.  For  the  product  of  two  binomials  having  a  common 
term  we  have  the  formula 

V.  The  square  of  the  polynomial  (a-\-h  —  G)  gives  the  formula 

(a  +  ft  _  c)2  =  a2  _|_  2,2  _j_  ^2  _|_  2  a&  -  2  ac  -  2  &c. 

VI.  The  cube  of  the  binomial  (a  +  I))  gives  the  formula 

{a  +  6)^  =  a^  +  3  a^h  +  3  fl&'  +  &'.  • 

Similarly,    {a  -  &)*  =  a*  -  3  a^b  +  Sab^  -  6^. 

ORAL   EXERCISES 

1.  Express  in  words  each  of  the  formulas  I  to  VI  which 
precede. 

Perform  the  indicated  operation : 

2.  (x  4-  3)2.  6.  {x''  -  xf.  10.  (3  x^  -  4  xcf. 

3.  (x-Bf.  7.  (2x-{-cy.  *  11.  (Tcc-f  4«xy. 

4.  (2x-{-  4)2.  8.  (x  -  3  c2)2.  12.  (x^  -  x-^. 

5.  (4.x -Sy.  9.  (4a^  +  a''^0'-  13.  (£c*-3x-'^)2. 

14.  (2  a^  -  a- ^)  (2  rt^  -  ft- ^)  (2  a^  -  a" ^)  -r-  (2  ^^  -  a"  ^). 

15.  (16£c2-24:r4-9)H-(4.'r-3)=?    Why? 

16.  (16  x^-i-S  xh  +  xV)  --  (4  £c  +  xc)  =  ?    Why  ? 


14  SECOND  COURSE  IX  ALGEBRA 

17.  (x  -  c)(x  +  c).  29.  (x  -h  3)(:r  +  4). 

18.  (x  -S)(x-\-  3).  30.  (x  -{-5)(x  +  7). 

19.  (x  -\-6)(x-  6).  31.  (a  +  S)(a  +  6). 

20.  ((f  -Sc)(a-\-S  c).  32.  (/>  -  3)  (^»  -  4). 

21.  (m  -  a;)(a;  +  7;t).         -  33.  (m  -  5)(m  -  10). 

22.  (4  a  +  c)  (c  -  4  «).  34.  (c  -  1)  (c  +  2). 

23.  (4:x  -Sc)(3c  +  4.X).  35.  (.r  -  S)(x  +  5). 

24.  (a^  -f  4  c)  (^2  -  4  r).  36.  (6'  -  7)  (S  +  4). 

25.  (aj«  -  ex) (x^  +  r^-).  37.  (x  -  12) (a?  +  3). 

26.  (4c»-a«)(««  +  4c«).  38.  (a  -  4c)(«^  +  2^). 

ic  +  m  ~^    Why?  ^^^^  (a«^- 2a-^)(a^  +  5a-=^). 

^„    rt*  —  16 c^  41.  (ax  —  ac)(ax  —  3 ac). 

28.  —5 i— =  ?    Whv^  ^  ^ 

a'  -  4c  ^  '  42.  (ca;  -  4c2)(ca^  -\-Sc^. 

a''  —  6a 
|4.  (a  +'^  4-  cf.        46.  («  -  c  +  a;)^.        48.  (a  -  c  +  2f. 
45.  (a  -f  c  +  xf.        47.  (a  -  c  -  ic)".        49.  (x  -  c  -  3  af. 


50.  ^^  +  9o-  +  a.^-6ac-f  2^0.-6...  =  .    Why  ^ 

a  —  3c-\-x       '  "^  ' 

51.  (2c-2a-4rr)2.  -54.  (4a  -  3c -^  2^;)^. 

52.  (3  c  -  5  «  +  2.r)2.  55.  (a  +  a'^  -  3y. 

53.  (x2«  +  cc«  -  5)"-^.  56.  («^  -  a-^  +  4)^. 

57.  Can  the  expression  in  Exercise  44   be  squared   as   a 
binomial  ?   Explain. 

58.  (x  +  ey.  62.  (x  +  2f.  66.  {x^  +  a)^ 

59.  [x  -  c)\  63.-  \x  -  2)».  67.  (a;^  -  2f. 

60.  (a^  -  If.  .  64.  (ic  +  3)«.  68.  (5  «  -  4  c)8. 

61.  lx  +  1)\  65.  (a-  -  4)».  69.  (2  .t^  -  7  ^-2)^ 

70.  (^8  -  6x^  +  12a;  -  8)--(a;  -  2)  =  ?   Why  ? 

71.  (8-12i  +  6a;2-a-8)--(4-4ic  +  ic2^=?    Why? 


FimDAMENTAL  OPERATIONS  15 

72.  Squai^  (5  —  7)  as  a  binomial  and  check  the  result  by 
subtracting  7  from  5  and  squaring  the  difference  obtained. 

73.  Square  x-\-9  and  —  ic  —  9.    Compare  results  and  explain. 

74.  Find   the   product  of  (9  -  4)  (9  +  3)   by  the  formula. 
Verify  by  simplifying  each  binomial  and  then  multiplying. 

76.  Square  :  (a)  42,  (b)  59,  (c)  73,  (d)  105,  (e)  97,  and  (/)  1005. 

76.  Expand  (4  +  9  —  5)^  by  "the  formula.    Verify  by  simpli- 
fying and  then  squaring. 

77.  Expand  (3  —  2)^  by  formula.    Verify  by  simplifying  the 
binomial  and  then  cubing  the  result. 

78.  Expand  (a  —  2  by  and  (2  b  —  rt)l    Compare  results  and 
explain. 

79.  What  must  be  added  to  9x^  -{-  6x  to  complete  the  tri- 
nomial square  ? 

80.  What  must  be  added  to  or  subtracted  from  16  a^  -f-  9  to 
complete  the  trinomial  square  ?    Why  ? 

Form  a  perfect  trinomial  square  of : 

81.  x^  -  ?  +  9.  90.  25cc2  -  12a^  -f  ? 

82.  4x2  + ?  +  l.  91.  a^-{-?  +  a-\ 

83.  4.x^-?  +  9a\  92.  a2x-?  +  a-2- 

84.  x'-^4^x-\-?  93.  a2x_9_j_i6a-2»^. 

85.  4  x'^  +  4  X  +  ?  94.  a'^^  +  10  +  ? 

86.  9x2±24x  +  ?  95.  a''-?  +  4.9a-\ 

87.  ?  ±  12x  +  9.'  ■      96.  ^6  -  6^2  +  ? 

88.  4x2-18«x  +  ?  97.  a*^-12a^  +  ? 

89.  9x2-4ax  +  ?  98.  4a«^ -f- ? -f- 25  a-^^. 


CHAPTER  II 

FACTORING 

11.  Definitions.  Factoring  is  the  process  of  finding  the  two  or 
more  expressions  whose  product  is  equal  to  a  given  expression. 

In  multiplication  we  have  two  factors  given  and  are  required  to 
find  their  product.  In  division  we  have  the  product  and  one  factor 
given  and  are  required  to  find  the  other  factor.  In  factoring,  how- 
ever, the  problem  is  a  little  more  difficult,  for  we  have  only  the 
product  given,  and  our  experience  is  supposed  to  enable  us  to  deter- 
mine the  factors. 

In  this  chapter  (except  in  §  17)  only  those  expressions  and 
factors  which  are  rational  will  be  considered. 

An  integral  expression  is  here  regarded  as  prime  when  no 
two  rational  integral  expressions  can  be  found  (except  the  ex- 
pression itself  and  1)  whose  product  is  the  given  expression. 

It  must  be  remembered  that  to  factor  an  integral  expression 
means  to  resolve  it  into  its  prime  factors. 

The  methods  of  this  chapter  enable  one  to  factor  all  integral, 
rational  expressions  in  one  letter  which  are  not  prime,  as  well  as 
some  of  the  simpler  expressions  in  two  letters.  No  attempt  is  made 
even  to  define  what  is  meant  by  prime  factors  of  expressions  which 
are  not  rational  and  integral. 

There  is  no  simple  operation  the  performance  of  which 
makes  us  sure  that  we  have  found  the  prime  factors  of  a  given 
expression.  Only  insight  and  experience  enable  us  to  find 
prime  factors  with  certainty. 

A  partial  check,  however,  that  may  be  applied  to  all  the 
exercises  in  factoring,  consists  in  actually  multiplying  together 
the  factors  that  have  been  found.  The  result  should  be  the 
original  expression. 

The  types  of  factorable  expressions  previously  considered 
will  be  reviewed  in  §§12,  13,  14,  15,  16,  18,  and  20. 

16 


FACTORING  17 

12.  Polynomials  with  a  common  monomial  factor.    The  type 
^^^•^^^^^  ab^ac-ad. 

Factoring,        ab  -\-  ac  —  ad  =■  a(h  -\-  c  —  d). 

EXERCISES 

Factor : 

1.4^  +  8.    •   4.  2c6/-46'2-2.  7.  £c2«-3x«  +  12x. 

2.  5  -  10  a".     5.  ah  -  ac"  -  4.  ac.         8.  ?/2«  -  6  ^/^  +  2  y^-\ 
Z.  ax  — lay.     6.  8 xy -{- 21  y^  —  15 y^    9.  5xy -^SOy(x^  +  xy). 

10.  (7a^-21ab-\-7a)-Uax. 

11.  14  a^x^  -  21  a'^x^m  -  49  a*a^y. 

12.  (3  c^  -  3  cd)  -  a  (45  c^  -  15  c^x). 

13.  2^2^  +  3 +  12r^-'^-16sr^+2-f  8 sr^+*. 

13.  Polynomials  which  may  be  factored  by  grouping  terms. 

The  type  form  is  ,         ,   t     ,   i. 

•^^  ax  +  ay  +  bx  +  by. 

Factoring,  ax  +  ay  -\-'bx  -\-hy  =  (ax  +  ay)  +  (bx  +  by) 

=  a(x  +  y)+b(x~+yy 
=  (x  +  y)(a  +  b). 

EXERCISES 

Separate  into  polynomial  factors  : 

1.  3(x  +  y)-^  a  (x  +  y).  3.  5x(a  —  b)  +  (b  —  a). 

2.  a(x~-3)-b(;x-3).  4.  2c(r  -  2s)- 5d(2 s -r). 

5.  4  a?  —  4  2/  +  to  —  &?/. 

Hint.    4  x  —  4:  y  -\- bx  —  hj  =  4:(x  —  y)+ b(x  —  y),  etc. 

6.  3  r.r  +  6  «c  +  8  ax  -\-  4.x\ 

7.  -6cc2  +  10a;-h21rr7;^-35m. 

8.  a'^-x^-a'^-i-2x^-2x-  2. 

9.  rs  -  2s  +  3  r  -  6  -  5rcc  +  10:r. 

10.  x^«  — 3.t2«  — a-^H-S  —  6£("*«  +  2£c5«. 

11.  .r3«--2  _f_  2x«  +  i  -  15i('2«-3  -  10  +  10.r2«-^ 


18  SECOND  COURSE  IX  ALGEBRA 

14.  Trinomials  which  are  perfect  squares.    The  type  form  is 

Factoring,  a^  ^  2  ab  -\- b"' =  (a  ±  hf. 

EXERCISES 

Separate  into  binomial  factors  : 

1.  a:^  +  6a:  +  9.         4.  ^b^  -  12^.  +  4.    ^7.  r/^  +  2  +  "--■ 

2.  a2-12a4-36.     5.  28 ic  +  49 ic"  +  4.     8.  <r-2  +  .^--. 

3.  ic^  +  ic  +  iv  6.  9  7-2 -I- 49 -42/-.      9.  «4-G  +  9<fc-*. 
^v   , ^     cc2        8a:,,,,  13.  »■*«  -  10  «a:2«  ^  25  a". 

11.  81  aV  -  36  a^te  +  4  Z-^.     15.  a^x  _^  4  ^,- a.  _  4 

12.  1  +  4  a*6«c"  -  4  a%«.         16.  «^-  -  2  w-  +  a"  =^- 

17.  (x  -  2)2+  14(a;  -  2)  +  49. 

18.  4(a  +  5)2-12^»(a  +  5)+9^.-l 

19.  (.^  -  bf^  -  18  a;  (a  -  bf  +  81  x^ 

15.  The  quadratic  trinomial.    The  type  form  is 

x"" -\- bx  ■{■  c, 
Eor  factoring  expressions  of  this  type  we  have  the 
Rule.    I'^ind  two  numbers  whose  algebraic  product  is  c  a7id 

whose  algebi'aic  sum  is  h. 

Write  for  the  factors,  two  binomials  which  hare  x  for  their 

common  term  and  the  numbers  just  obtained  for  the  other  tei'ms. 

EXERCISES 
Separate  into  binomial  factors  : 

1.  a:2-8ic  +  12. 

Solution  :  The  two  numbers  whose  sum  is  —  8  and  whose  product 
is  +  12  are  -  2  and  -  6.    Therefore  a:^  -  8 a:  +  12  =  (.r  -2){x-  0). 

2.  x'-^x^  18.       5.  a^  -  12  -  11  a.       8.  x^  _  8a;  -  9. 

3.  x''  +  2x-  24.       6.  y  +  .3 a  -  .1.         9.  / V  +  6 rs  -  40. 

4.  x^  +  a:  -  f .  7.  c"  -  ac  -  90  a\     10.  <r  -\- 5  ^  Q>  a-\ 


FACTORING  19 

11.  1  -  9  a-^  4-  8  iPy.  15.  12  a*  -  aH  -  x\ 

12.  11  iiy-  12  d%^  +  ^•''.  16.  a'"  -20  +  19  «". 

13.  15m-^-14wi«-a--2.  17.  «;'  +  12a-2  +  7. 
Hint.    This  may  be  written  as  18.   (X^^  —  8  a" ^^  —  2. 

-  1  (x2  +  14  ^m  ^  15  m%  etc.  ^9    j^20  +  7  m"  -  m^". 

14.  90  +  X  -  i^'2.  20.  a^^  -  «^  -  6  a- 2^. 

21.  (»?  4-  ny  -  ^{m  +  7i)'"  -  22. 

16.  The  general  quadratic  trinomial.    The  type  form  is 
ax^  -\-bx-\-  c. 

This  important  type  really  includes  the  two  preceding  types. 

If  a  trinomial  of  this  type  has  two  rational  factors,  they 
have  the  forms  dx  +  e  and  fx  -\-  g. 

Now       {dx  +  e)  (fx  +  J/)  =  dfx^  +  fex  +  dgx  +  ge  (1) 

=  dfx'-h(fe  +  dg)x  +  ge.  (2) 

In  (2)  the  product  of  the  coefficient  of  x^,  df,  and  the  con- 
stant term,  ge,  is  dfge.  But  dfge  equals  fe  times  'c?^,  and  /e 
plus  dg  equals  the  coefficient  of  x.  Therefore,  if  ax^  -\-  bx  -{-  c 
has  rational  factors,  it  can  be  written  in  the  form  (1)  and  fac- 
tored by  grouping  terms.    Hence  the 

KuLE.  Find  two  numbers  whose  algebraic  product  is  ac  and 
whose  algebraic  sum  is  b. 

Replace  bx  by  two  terms  in  x  whose  respective  coefficients  are 
the  numbers  just  found,  and  factor  by  grouping  terms. 

EXERCISES 

Separate  into  binomial  factors  : 

1.  4a-2-7./-15. 

Solution  :  Here  ac  =  4  •(—  15),  or  —  60,  and  b  ——7.  The  numbers 
whose  product  is  —  60  and  whose  sum  is  —  7  are  -h  5  and  —  12. 
Hence  4  x-  —  7  a;  —  15  =  4  x^  —  12  a,  +  5  a:  —  15 

=  4  a:(x  -  3)  +  5 (x  -  3)  =  (a,-  -  3)(4  x  +  5). 

2.  2«2  -  3«  -  2.        4.  4«2  H-  a  -  5.  6.  5 r^  -  22r  -f  8. 

3.  3«^  +  8(t-3.        5.  9c2-71c-8.       7.  7  j:"' -f- 62  x  -  9. 


\ 


20  SECOND  COURSE  IN  ALGEBRA 

8.  6x^-hl9x-T.  14.  -Sn*-\-3n'-S. 

9.  Gx'-hlSx-d.  15.  6x^^-13x^  +  6. 

10.  2x^-\-7x-  16.  16.  20^2  -9xi/-  20/. 

11.  Sx^-ax-2a^  17.  2x^ -(a -I- 2b)x -j- af,. 

12.  4  a*  -  12  a^  _|_  9.  is.  5  m^"-^  +  9  am^-^  -  2  ^r. 

13.  25 -^  A c^P- 20 cd.  19.  6.t*« +  (3  -  2/)a;2«  - /. 

20.  20^2^.4-2-- 9rt-20^'2^-^ 

21.  6a2x+6_25a^  +  3^,!/-i4.4^2j,-2_ 

17.  The  factors  of  ax^ -\- bx  +  c  hy  formula.*  The  factors  of 
any  trinomial  of  this  type,  whether  they  are  rational  or  irra- 
tional, can  always  be  obtained  as  follows : 

Let  ax^-{-hx  +  c  =  0.  (1) 

Then  x^-^-x  +  -  =  0.  (2) 

a         a  ^  ^ 

If  (2)  be  solved  by  completing  the  square  (see  Exercise  1,  page  271, 

"  First  Course  in  Algebra  "),  we  obtain 

J.  n/' 

From  (3),  x "  '    ;" ^^  =  0,  (4) 


and  X =  0.  (6) 

2  a  ^  ^ 

From  (4)  and  (5), 

/    _  -h-h-\/b^-4:  ac\  (_-h-  -s/y^  -  4  ac 


X 

-h±^ 

-4ac 

2  a 

_   /,  +  V/>2  _ 

4  ac 

2  a 

_  h  _  V62  _ 

4  ac 

2a 


(6) 


Now  (6)  and  (2)  are  equivalent,  that  is,  they  have  the  same  roots. 
Further,  if  both  members  of  (2)  be  multiplied  by  a,  the  result  is  (1). 
If  both  members  of  (6)  be  multiplied  by  a,  we  obtain  an  equa- 
tion which  is  identical  with  (1),  as  can  be  verified  by  performing 
the  indicated  multiplication.    That  is, 

ax^  -\-bx-\-c 

/         _&^_V&2_4aA/        _&_  V62_4aA 

*  See  also  page  240. 


FACTORING  21 

EXAMPLE 
Factor  10  a;2- 7  a; -12. 
Here  a  =  10,  h  =—7,  and  c  =—12.   Using  (7)  as  a  formula  gives : 


2-2  -  7  a:  -  12 

-10  (x      -(-7)+V(-7)2-4.1()(-12)\/        ' 

7-V49+480' 

^^r                              20                         A 
=  2  (a;  -  1)  5  (x-  +  1)  =  (2  a:  -  3)  (5  X  +  4). 

20 

We  may  infer  from  the  preceding  work  that  a  trinomial  of  the 
form  ax^  -{■  bx  +  c  has  rational  factors  only  when  ^^  —  4  ac  is  a  perfect 
square,  . 

EXERCISES 

Factor  by  use  of  the  formula  : 

1.  Sx^  +  5x  +  2.  10.  x^-6x-^l. 

2.  Qx^  +  x-2.  11.  2x^-\-5x  +  l. 

3.  x^  +  4.x-j-l.  12.  3x^  +  6x-l. 

4.  Sx''-5x-12.  13.  4x2-8x  +  3. 

5.  5x^  -  19x  -  4.       /  14.  5a;2  +  7  a;  -  3. 

6.  7£c2  -  20a;  -  3.  15.  x'-2x-{-l-n. 

7.  15ic2- 11^-14.  16.  x2  + 6a; -47^  +  9. 

8.  2>x'  +  ^mx-2m\  17.  ic^ -(t^  +  5)a;  +  (2^1  +  6). 

9.  4  a;2  —  5  Tia;  —  6  72-1  1^.  x^  —  nx  —  2  x  -\-  ^  n  —  3. 

18.  A  binomial  the  difference  of  two  squares.    The  type  form  is 

Factoring,  a^  —  jj'^  =  (a -\- h)  (a  —  b). 

More  generally, 

a'  +2ab  +  b^-c^+2cd-d:' 

=  a^+2ab  +  b-'-(c^-2cd-{-cF) 

=  ((l  +  bf-(G-df 

■       '  =(a  +  b  +  c  -  d)(a  +  b  -  c  +  d). 


Factor : 

1.  m''  -  n\ 

6. 

2^x^-\Sih\ 

2.  «^  -  c\ 

7. 

(v'-a-^. 

3.  x"-  -  4. 

8. 

1-9.*V. 

^.x--\. 
5.  81 -.T*.  . 

9. 

,      4c^ 
^^        25' 

22  JSECU.ND  CUUUJSE  IN  ALCiKBKA 

EXERCISES 

10.  16r/«-25^^^/i^ 

11.  «2x_^-2a:^ 

12.  a^-a-\ 

13.  (a4-^')'-l- 

14.  (a  -  xf  -  4. 

15.  9-(2  +  a;)2.  21.  r<2  ^  2  «:«  +  ic^  _  9. 

16.  16-(a;-a)««.  Hint,   a^  +  2  «a.- +  x^  -  9 

17.  52«  _  (,,,  _  i)ioa  =  («  +  ^)'  -  ^- 

18.  (,,  +  cf  -  (m  +  nf.  22.  a^  -  4  ac  +  4  c^  -  x\ 

19.  (a  +  cf  -  (m  -  /i)l  23.  25  - 10  a;  +  x^  -  16  yy^l 

20.  (a  -  ir)2'-  _  (c  -  5)2.  24.  9  -  12  «  +  4  r^.2  _  h^^\ 

25.  7?i2  _  ^^2  -  6  a  -  9. 
Hint,    m^  -a'^-Qa-^  =  m^  -  (a^  +  6  a  +  9)  =  ?«2  _  (^^^  +  3>)2_ 

26.  .T^  -4:f  4-  20.?/  -  25.         28.  4  m*  +  30  A  -  9£c2  _  25n^ 

27.  l&-^tv'-\-12ax-4.x\     29.  -  28  c^c^-^  -  49  c«  +  1  -  4  </•». 

30.  12 /•«  -  36  +  52«  - /•2^ 

31.  (m-2)2-4?^2  4-28  7l-49. 

32.  a-2-6(r  +  9-/  +  8ay-16</l 

33.  4&<^  4- 40^-4^ -46-^*2 +  1. 

34.  6c  +  7^2  _  1  _  9^2  -  47Jj  +  4H 

35.  A2_4y2_i0A  +  8a'.y  +  25-4a'2. 

36.  2^x^  -  20x  +  4.  -  4.1/  -  ^ cv"  -12 ay. 

19.  Expressions  reducible  to  the  difference  of  two  squares.   The 

type  form  is 

If  k  has  such  a  value  that  the  trinomial  is  not  a  perfect  square^ 
a  trinomial  of  this  type  can  often  be  written  as  the  difference  of 
two  squares.  Thus,  if  Zr  =1,  the  adding  and  subtracting  of  a%^ 
transforms  the  expression  into  the  difference  of  two  squares. 


FACTORING  23 

EXAMPLES 

1.  Factor  a""  +  a%''  +  h\ 

Solution  :  a^  +  a%''  +  ?*'  =  «*  +  2  a%'^  +  h^  -  aHP-  "* 

=  (a2  +  6^  +  a6)  (a^  +  V^  -  ah). 

2.  Factor  49  It"  +  34  hVc"  +  25  /v^ 

Solution :    If  36  h'^k'^  is  added,  the  expression  becomes  a  perfect 
trinomial  square.    Adding  and  subtracting  36  li^k^,  we  have 

49  /i^  +  34  AU-2  +  25  k^  =  49  /^^  +  70  A%2  +  25  k^  -  36  /<2X,-2 
=  (7  A2  +  5  1-2)2  _  (6  hkf 
=  (7  A2  +  5  ^■2  +  6  Z^-)  (7  A2  +  5  F  -  6  7<^-). 

EXERCISES 

F'actor : 

.1.  a-*  +  ic'  +  l.  12-  ic'  +  .r'  +  l. 

3.  X*  +  4a:2  -f- 16.  14.  16  +  4x*  +  icl 

4.  16/4-4?/^  +  l.  15.  25v/^-ll/4-l- 

5.  c*  +  c^tZ^  +  25 1^^  16.  f  +  16  ?/'  +  256. 

6.  1-19/ +  25?/^.  17.  ^x^y  +  ^xY  +  ^xif. 

7.  4  a-^  +  3  a:y  +  9  //.  18.  16  h"  -  33  h^k^  +  36  k\ 

8.  4  ic'^  -  28  xY  4-  9  7/\  19.  25  0*  -  51  c^d'  +  49  6^*. 
*      9.  9  c'  -  55  (?d?  +  25  d\  20.  49  6t*  -  32  a^"-  +  64  ^'*. 

10.  9^«-19a^^»2_j_25^,4  21.  64  cc*  +  119  it-y  +  8iy. 

11.  49  A^  -  44 /i^yt*  +  4  7vl  22.  81  «^  -  171  a^Z*'^  +  25  Z»^ 

23.  1  +  4  x^.      Hint.    1  +  4  x*  =  1  +  4  a:2  +  4  a;4  -  4  a:2. 

24.  64  c'^  +  1.  26.  ic«  +  4  t/I  28.  cc*"  +  4  ?/««. 

25.  cr*  +  4  y^  27.  a-^  +  64.  29.  a'^c^'^  +  64  6*^  +  -^. 

20.  The  sum  or  difference  of  two  cubes.    The  type  form  is 

a'  ±  &^ 

Factoring,       a»  +  Z*^  =  (^  +  h)  {a^  -  ah  ^-  V^, 
a^  -h^  =  (a  -  b)  («.2  +  ah  +  ^.2). 


EXERCISES 


Factor : 

1.  ar«  -f  64. 

Solution  :  ^r^  +  64  = 

a;8  +  48  =  (x  +  4)  (a:2  -  x .  4  +  42) 
=  (x  +  4)(x2-4a:  +  16). 

2.  x^  4-  27. 

3.  a'  -  64. 

4.  8  +  wi». 

o      8      «*                      9.  1- 

7.  x8-7/-l               ^^"  27 

5.  27-7ri«. 

8.  8a^  +  27^'^.           11.  a-«- 

-/. 

12.  a:«  +  7/^ 

16.  x'^-Sx^-j-^x- 

-28. 

13.  x^  -  a\ 

17.  a8^.V-8fP. 

14.  a:«  4-  a'. 

18.  a;8«-7/-8^ 

15.  (a:  +  yf  - 

-  8.                19.  c««  +  27  d^\ 

21.  The  Remainder  Theorem.  If  any  rational  integral  expres- 
sion in  X  be  divided  by  x  —  n,  the  remainder  is  the  same  as 
the  original  expression  with  n  substituted  for  x. 


EXAMPLE 


c^-5i:    +6 

p2  —  « r 


x  +  (n  —  5) 


(n  -  5)  a;  +  6 

(w  —  5)  X  —  n^  +  5  n  , 

ri^  _  5  ,^  _j_  (j  ^  Remainder 

Here  the  remainder  n^  —  5n  +  6  is  the  same  as  iK^  —  5 a;  +  6 
when  n  is  substituted  for  x. 

Now  if  71  is  a  letter  or  a  number  such  that  the  remainder 
n^  —  5  71  +  6  is  zero,  the  division  is  exact ;  and  the  value  of  n, 
if  substituted  for  x,  will  make  x"^  —  5x  -\-  6  zero  also. 

Hence,  if  by  trial  we  can  discover  a  number  n  which,  when 
put  for  X,  makes  x^  —  5x  -\-  6  zero,  x  —  n  will  be  an  exact  di- 
visor of  «2  —  5  £c  -f  6.  If  2  is  put  for  cc  in  cc^  —  5  a;  +  6,  we  get 
4  — 10  +  6,  or  zero.   Therefore  ic  —  2  is  a  factor  of  x^  —  5x  +  &. 

The  last  paragraph  illustrates  the  following  theorem  : 


FACTORING  25 

22.  Factor  Theorem.  If  any  rational  integral  expression  in  x 
becomes  zero  when  any  number  n  is  put  for  x,  x  —  n  is  a  factor 
of  the  expression. 

The  Factor  Theorem  may  be  used  to  factor  many  of  the 
preceding  exercises.  Moreover,  many  expressions  which,  by 
previous  methods,  are  very  difficult  to  factor,  may  be  readily 
factored  by  the  aid  of  this  theorem. 

Note.  By  means  of  the  Factor  Theorem  we  are  able  to  solve  cubic 
equations  when  the  roots  are  integers.  The  solution  of  the  general 
cubic  equation  is  one  of  the  famous  problems  of  mathematics,  and 
one  which  is  accompanied  by  many  interesting  applications.  This 
problem  was  first  solved  by  the  Italian,  Tartaglia,  about  1530,  but 
was  published  by  Cardan,  to  whom  Tartaglia  explained  his  solution 
on  the  pledge  that  he  would  not  divulge  it.  For  many  years  the  credit 
for  the  discovery  was  given  to  Cardan,  and  to  this  day  it  is  usually 
called  Cardan's  solution. 


m 


EXAMPLES 
actor : 


-  1.  x^'  +  x-  2. 

Solution :  If  a;  —  w  is  a  factor  of  x^  -\-  x  —  2,  then  n  must  be  an 
integral  divisor  of  2.  Now  the  integral  divisors  of  2  are  + 1, 
- 1,  +2,  and  '-  2.  If  1  be  put  for  x,  x^  +  x  -  2  =1  +  1-2  =  0. 
Therefore  x  —  1  is  a  factor  of  r^  -\-  x  —  2.  Dividing  x^  +  x  —  2  by 
,r  —  1,  the  quotient  is  a'^  +  a;  +  2.  None  of  the  integral  divisors  of  2, 
when  put  for  x,  make  x^  -\-  x  +  2  zero ;  hence  a:^  +  a;  +  2  is  prime. 

Therefore  x^  +  x  -  2  =  (x -l)(x^ -\-  x  +  2). 

2.  ^^^2x''-5x~6. 

Solution :  The  integral  divisors  of  6  are  +1,  —  1,  +2,-2,  +3, 

-  3,  +  6,  and  -  6.    If  we  put  1  for  x,  a;^  +  2  a;^  -  5  a;  -  6  =  1  +  2  - 

5  -  6  =  -  8.    If  we  put  -  1  for  X,  a:^  +  2  a;2  -  5  a:  -  6  =r  _  1  +  2  +  5  - 

6  =  0.    Therefore  ar  —  (—  1)  or  a:  +1  is  a  factor.   Dividing  a:^  +  2  x^  — 
5  a;  —  6  by  X  +  1,  the  quotient  is  a:^  +  x  —  6,  which  equals  (x  +  3)(a:  —  2). 

Therefore  a:^  +  2  a:2  -  5  a:  -  6  =  (a;  +  l)(a:  -  2)(a:  +  3). 

In  the  following  exercises,  when  searching  for  the  values  of 
X  which  will  make  the  given  expression  zero,  only  integral  divi- 
sors of  the  last  term  of  the  expression  (arranged  according  to 
the  descending  powers  of  x)  need  be  tried. 


2t>  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

1.  Divide  x^  -{-  bx  -{-  c.hy  x  —  n  and  show  that  the  remainder 
is  n^  -\-hn  -\-  c. 

2.  Find  the  remainder  in  (x^  +  ax"^  -\- hx  -{-  c) -^  (x  —  71). 
Find,  by  the  Remainder  Theorem,  the  remainder  when : 

3.  ic^  —  ic  —  8  is  divided  hj  x  —  S. 

4.  x*  —  X  -\-  6  is  divided  by  a?  +  2  by  synthetic  division. 

Factor : 

(Instead  of  ordinary  substitution,  the  student  should  use  synthetic 
division  to  find  whether  the  remainder  is  or  is  not  zero.) 

5.  a-8-3a;  +  2.^  8.  a;»  -  ic  -  6.  11.  x^-llx-6. 
-     e.  x^-4x-\-3.           9.x^-x  +  6.              12.  :e'-14ir-8. 

7.  x^  +  2x-h3.         10.  .T^-lla:  +  6.  13.  ;r'-27ic-10. 

14.  x^  +  Sx^  +  Sx-\-2/:\     18.  Ax^-Sx-l. 

15.  x^  +  4:x^  +  5a;  +  2.         19.  x^  -5a^x  ■j-2aK 

16.  x^  -6x''-\-llx-  6.       20.  x^  -  Itiv'x  -  6ml 

17.  ic^  -  11  a;'  +  2  X  +  12.      21.  x^  -  2  tix'' -  6  n^x -\- Q>  n\ 

23.  The  sum  or  difference  of  two  like  powers.    The  tyj^e  form  is 
a"  ±  &«. 

The  cases  in  which  a"  ±  If"  is  divisible  by  a  +  ^>  or  a  —  ^^  can 
be  determined  by  the  Factor  Theorem. 

Thus  in  a"  —  Z*",  n  being  either  an  odd  or  an  even  integer, 
let  a  =  b.  Then  a"  —  If*  becomes  h''  —  J/'^  0.  Therefore  a  —  b 
is  a  factor  of  a"  —  ^''. 

In  a"  —  ^",  71  being  even,  let  a  =  -\-  h  or  —  b.  Then  a"  —  //' 
becomes  h^  —  h^  =  0,  since  both  (+  b)"  and  (—  by  are  positive 
when  71  is  even.  Therefore  when  n  is  even,  both  a  —  Z*  and  a  -\-b 
are  exact  divisors  of  a"  —  Z*". 

In  a*  +  ^"j  71  being  even,  let  a  equal  either  -f-  Z»  or  —  b.  Then 
a"  -f  h^  becomes  Z/"  +  ^",  which  is  not  zero.  Therefore  a"  +  &'*  is 
never  divisible  hj  a  -{-  b  ov  a  —  b  when  n  is  even. 


FACTORmG  27 

In  a"  +  h"",  n  being  odd,  let  a  =  —  h.  Then  «"  +  &**  becomes 
(—  by  4-  Z»"  =  0,  since  (—  by  is  negative  when  ?i  is  oc?c?.  There- 
fore when  n  is  odd,  a  +  Z»  is  a  divisor  of  a"  +  Z»". 

Summing  up : 

I.  a""  —  J"  is  always  divisible  hj  a  —  b. 
II.  a"  —  ^",  when  ti  is  e?;e?i,  is  always  divisible  by  a  —  b 
and  a  -\-b. 

III.  a"  +  Z*"  is  never  divisible  by  a  —  ^. 

IV.  a"  +  J",  when  n  is  otZc?,  is  always  divisible  by  a  +  /^. 

It  is  worth  noting  that  «/*  +  Z*"  is  usually  prime  when  n  is  a 
power  of  2.    (See,  however,  Exercises  23-29,  page  23.) 
Thus  a^  4-  ^^,  a''  +  b'^,  a^  4-  ^^,  etc.,  are  prime. 
In  every  other  case  a""  +  i"  is  not  prime. 
Thus        ^  a^  +  b^  =  (ay  +  {by, 

a^^  +  b^''  =  lay  +  (by,QtQ. 


EXAMPLES 

1.  Factor  a^  -  b\ 

Solution :  a^  -  //  =  (a  -  ft)  (a^  +  ««/>  +  a2^2  ^  ^^/^a  +  j4^^ 

2.  Factor  a'  -  b\ 

Solution  :  a^  —  h^  is  divisible  hy  a  +  b  and  a  —  h.  It  is  better,  how- 
ever, to  regard  all  such  binomials  with  even  exponents  as  the  differ- 
ence of  two  squares.    Thus 

rt^  -  ft6  ^  (a3  _  i^^  (a3  +  j3>)^  etc. 

1^^.     3.  Factor  a^  -\-  b^. 

Solution  :  a^  +  ft^  ^  («  +  h)  {a^  -  a^/,  +  anP-  -  «ft3  +  j^Ay 

Note  the  signs  of  the  second  factor  in  Examples  1  and  3,  —  all  plus 
in  one  case,  alternately  plus  and  minus  in  the  other. 

4.  Factor  x^  —  i/^. 
Solution:  x^  —  i/^  =  x'^  -  {y^ 

=  (X  - ,/)  [x^  +  x'  (fy  +  x^  (fy  +  X  (fy  +  (yy^ 

=  (x-  if)  {x'^  +  xhf  +  xhf  +  xif  +?/«). 


2«  SECOND  COURSE  IN  ALUEBKA 

5.  Factor  32  a-^'^  +  y^^ 

Solution:  32 a;" +  3/" 

=  (2x»)6  +  (y8)6 

=  (2  a;8  +  y8)  [(2  ar«)*  -  (2  ar^)*  (y*) 

+  (2  x^y^  (yy  -  (2  a-8)  (fy  +  (y»)*  ] 
=  (2  a;8  +  y^)  (10  a-12  _  8  xY  +  4  x«y«  -  2  a:V  +  yi2-), 

EXERCISES 


Factor : 

1. 

iz;^  +  l. 

2. 

x'-[-f. 

<3. 

x'-l. 

4. 

x'-y'r 

5. 

x'  +  l. 

'6. 

x'  -  32. 

7. 

243  +  a'. 

8. 

3125  -  c\ 

15. 

x^'  -  a\ 

9. 

c'  - 128. 

16. 

x''-\-a}\ 

10. 

c'  -  x^\ 

17. 

32ic«-l. 

11. 

x^'  +  ^. 

18. 

x'-^2y^\ 

12. 

1  +  c'd}\ 

19. 

128a^^-l. 

13. 

x''  +  y\ 

20. 

243£c^-M. 

14. 

c'l'  +  128. 

21. 

1024- 243  a;^ 

24.  General  directions  for  factoring.  The  following  suggestions 
will  prove  helpful : 

I.  First  look  for  a  common  monomial  factor,  and  if  there  is 
one  (other  than  1),  separate  the  expression  into  its  greatest 
monomial  factor  and  the  corresponding  polynomial  factor. 

II.  Then  determine,  by  the  form  of  the  polynomial  factor,  with 
tvhich  of  the  following  types  it  should  he  classed,  and  iise  the 
methods  of  factoring  applicable  to  that  type. 

1.  ax+ay-\-hx-\-hy,      5.  a" ±6". 

2.  a'  +  2a6+&^  ^a'-h'', 

3.  x"  +  bx-\-c.  6.  j  a'  +  2  a&  +  &'  -  c^ 

4.  ax^  +  bx+c,  [a'  ^2  ah  ^h''  -  c""  -2  cd-  d\ 

7.  a*  +  /fa2&2  +  &*. 

III.  Proceed  again  as  in  II  ivith  each  polynomial  factor 
obtained  until  the  original  expression  has  been  separated  into 
its  prime  factors. 

lY:  If  the  preceding  steps  fail,  try  the  Factor  Theorem. 


FACTORING  29 

REVIEW  EXERCISES 
Factor : 

1.  x^  —  X.  ,,2^1 

11.  ^'  -  ^  -  ^  • 

2.  x'-x.  ^6 

3.  x''  -  x\*^  12.  16x'-\-Sx^-  3. 

4.  X''*  —  2  x^  +  1.  13.  a^  —  a  +  a^^  —  ^. 

5.  ic^  —  2  X*  +  1.  1^-  3  x^  —  15  x^  +  12. 

6.  x'-Sx'-{-16x.  15.  x^-x2-4x  +  4. 

7.  x*-10x2  +  9.  16.  47i«  + 4871^-2871*. 

8.  X*  -  13  x^  +  36.  17.  12  a  -  39  mj  -  51  af. 

9.  3«2x*-12aV  +  12a\  18.  x*  -  3x^  +  4^^  -  12x. 
10.  18aV-24a2x-10al  19.  3^^^  +  3a' -  27a  -  27. 

20.  2^^^*  +  3  a%  -Sab -12b. 

21.  4a2-a*  +  81  +  10a2ic-36a-25x2. 

22.  12cd^  -  6a^x  -  a'  -h  4.c^  +  9d'  -  9x\ 


23.  x«  + 1. 

29. 

x'- 

-64x«. 

36.  xi2  +  64  7/^. 

24..^       16- 

30. 
31. 

x'^ 
x"^ 

-64. 

37.  32x^0  +  /°.-' 

38.  x^^  -  i/\ 

25.  x''  -  if. 

32. 

x^' 

+  V. 

39.  16xi«-l. 

26.  x'  -  y\ 

33. 

x^ 

-8. 

40.  x^  +  y^\ 

27.  x^-x'. 

34. 

x^- 

-y\ 

41.  243  -  A^ 

28.  x^2  +  7/^2. 

35. 

64 

x^^-4.x\ 

42.  5-*^-l. 

43.  20-x-icl 

50. 

a^- 

.  ^2  +  a  +  1.^ 

44.  10-10c"6Z*. 

51. 

a^- 

.  ^3  ^  ^  _  1 

45.  2cd-c^-d' 

52. 

a«- 

-  a*  —  a^  +  a. 

46.  £c^  -  X*  -  x^  +  xl 

53. 

hd^ 

-bcd-l^c^ 

47.  £c*-9a;=^-x 

+  3. 

54. 

x^- 

.  3a;2  +  8x-12. 

48.  x''-lxhf-\- 

81/ 

55. 

121 

x*-476xy+100//* 

49.  4c*  +  20c^(/- 

-11, 

i?(]?. 

56. 

x'- 

-a-2  +  12x7/-36  7/2. 

57.  /  - 

-187/^  + 

81- 

-16x*- 

-24xy-97/«. 

'do  SECOND  COURSE  IX  ALGEBRA 

58.  4  h'  +  32  A^A;'.  61.  U  +  5x'-  x^\ 

59.  W^-1024A;^  62.  1- x -{- x""  -  x'. 

60.  x^  -  83 x^  -f-  289 x\  63.  x^-y'-x-y. 

64.  x^  —  if  ■\-  X  —  y. 

65.  289-100a2_^,2_20«Z,. 

66.  625  a«  -  169  e/^  +  78  cd''  -  9  c^. 
67.  c'-2187. 

.>A«8.  a;««-125/». 

69.  5  —  8  a;  -  4  ccl 

70.  4a;*-37ic2  +  9. 

71.  256-167.-^4-8A2A;2-A*. 

77.  ii^  -  9  ^^  -  8  a/>  +  6  cr/ 

78.  4A-«-20A-Vc  +  25/r 

79.  4  a;*  —  9  x^  -  9. 

80.  a;2«  -  2  ^'^  -  15. 

81.  a'  -h  «  +  ^''  +  ^>. 

82.  a;2«-12a;"-f-36. 

83.  25a^2c^5Q^c_39 

89.  a;^  +  3ic2 +9:^  +  27. 

90.  ic»_6x2  +  12a;-7. 

,  91.  4  a^Z'^  -  40  a%\?-  4- 100  Z'V. 

92.  4.T*-25/  +  10/-12^2_^8. 

93.  x^  -  k^  4-  6^;?/  +  6«  +  9?/2  -  9A*. 

94.  e'^  -  2  4-  e-2^.  99.  a^^  -  7  ic/  +  6  /. 

95.  e2a._5^6e-2^.  100.  e'^  +  ^'^ -f- e^  +  1. 

96.  3ic2a^5a,a_28.  101.  ff^^-^  -  10  4- 25^2-4*. 

97.  e^^  -^  2  e^  -  24  e"^.  102.  ^'^  — 6"^^  4-  3  e"^  --  3  e^. 

98.  6e^*-5e-2-^-13.  103.  ^^+8  4- ^^+2  _  g3-x_  ^2- 

104.  .T^y-  +  xz^  4-  .T^  4-  .T^rv  +  yz^  4-  ?/^-~  4-  2  x?/«. 

105.  a^*«  -  a^^  4- «c«  -  ^8^  4- ^*c-5  -  ^.^c. 


72. 

x^4-4. 

73. 

x^''  -  729. 

74. 

a4x  +  8^e4 

75. 

^t*  +  225  a-^ 

-39. 

76. 

x^-^x^J^Vlx- 

-c^ 

4-16^*2 

—  1 

6a/»-2_9a2 

-^>-*. 

84. 

64  4-  6'1 

85. 

128  -  a^^l 

86. 

3a;'4-10.T- 

-8. 

87. 

oJ>  —  «3  _  «2 

4-1. 

88. 

26'2  +  3«Z- 

-21d\ 

FACTORING  31 

25.  Solution  of  equations  by  factoring.  For  solving  an  equa- 
tion by  factoring  we  have  the 

Rule.  Transpose  the  terms  so  that  the  right-hand  member  is 
zero.  Then  factor  the  member  on  the  left,  set  each  factor  which 
contains  an  unknown  equal  to  zero,  and  solve  the  resulting 
equations. 

One  should  not  divide  each  member  of  an  equation  by  an 
expression  containing  the  unknown,  for  by  this  process  one  or 
more  roots  are  usually  lost  (see  Example  3,  page  44). 

EXERCISES 

Solve  by  factoring  and  check  :  • 

1.  0^2  -  25  =  0.  '6.  X*  -f  4  =  5ic'. 

2.  x^  +  10=lx.  7.  f=l^t^-^&t\ 

3.  2/2  -  9y  =  0.  8.  3ic2  -  xh  -  2//  =  0, 

4.  ^.2  _  ^^  ^  30^2^  9    x^-2x'  =  x-2. 
5.4^^-36^  =  0.                           10.  a;'"- 2af +  1=0. 

11.  x2^-8a;-  +  16  =  0. 

12.  x^  +  bx''-l^x-12  =  0. 
Hint.    Apply  the  Factor  Theorem. 

13.  x^-Qx''  +  12x  =  S. 

14.  x^  +  3a;^-8a;2  +  16  =  12a-. 

/ 15.  x^  +  Qa^  =  2  ax''  +  5  a'x.  ' 

16.  ic^  +  9  ax^  =  9  £c*a  4-  ^^^ 

17.  x^  ^bxh  -  l^xc"  -  80c«  =  0. 

26.  Highest  common  factor.  One  method  of  finding  the  H.C.F. 
of  two  or  more  rational,  integral  expressions,  which  can  be 
readily  factored,  is  stated  in  the 

Rule.  Separate  each  expression  into  its  prime  factors.  Then 
find  the  product  of  such  factors  as  occur  in  each  expres- 
sion, using  each  factor  the  least  number  of  times  it  occurs  in 
any  one  expression. 


S'A  8KCUND  CULK8K  lis'   ALGEJBltA 

EXERCISES 
Find  the  H.C.F.  of  the  following : 

1.  28,  56,  84,  and  35. 

2.  225,  120,  210,  and  135. 

3.  198,  495,  693,  and  1155. 

4.  816,  1224,  1360,  and  4080. 

5.  91a;y,  133  a;y,  and  343  ^y. 

6.  0-^^-9,  x'^-{-2x-S,  and  x'-\-Sx. 

7.  a^  _  9a  4- 14,  a^  -  4,  and  5a^-  10a. 

8.  ic»  +  27,  2x2  +  3a;  -  g^  ^^^  5^8  _|_  ^^^2 

9.  4:x^  H-  20ic,  ic»  +  4ic2  -5x,  and  4:ax^  +  20rtic. 

10.  2x^-\-Sx,3ax^+6ax-^6  ax%  and  3  ax'  +  10  ax^ 

11.  32  —  x^,  x^y  —  4:y,  Sxy  —  6y,  and  a;^  —  4  cc  +  4. 

12.  l(x  +  tj)(x-  y)y,  x*-2  xV  +  2/^  and  (a;«  -  y'f. 

13.  ic^  +  a'^  x^  —  a^%  x^  —  Sxa'  —  4:  a^%  and  x^  +  «*"• 

14.  a^'  +  4.a'b\  a'^'  +  H'' -\- 2 aV"  -  2a'  +  H\  d^ndi  a'' -1QJ^\ 

27.  Lowest  common  multiple.  The  lowest  common  multiple  of 
two  or  more  rational,  integral  expressions  is  the  expression 
of  lowest  degree  which  will  exactly  contain  each. 

For  such  expressions  as  can  be  readily  factored  the  method 
of  finding  the  L.C.M.  is  stated  in  the 

Rule.  Separate  each  expression  into  its  prime  factors.  Then 
find  the  product  of  all  the  different  prime  foAitors,  using  each 
factor  the  greatest  number  of  times  it  occurs  in  any  one  expression. 

EXAMPLE 

Find  the  L.C.M.  of 
12  icV  +  12  xy\  4  cc«  -  4  xy%  and  %x^y-%  x^  +  8  x\f. 

Solution  :  12  x^y  +  12  a:?/*  =  2^ .  3  •  xy  {x  +  ?/)  (ar^  -  xy  ■\-  y^). 
4  a;^  —  4  xy^  =  2^ -  x (x  +  y)(x  —  y). 
8xhj-S  xy  +  8  xy  =  28 x^y  (x^  -  xy  ■\-  y^). 
Therefore  the  L.C.M.  is  2^  •  3  xhj{x  +  y){x  -  y) (x^  -  xy  ^-  tf). 


FACTORING  33 

EXERCISES 

Find  the  L.C.M.  of: 

1.  24,  30,  and  54. 

2.  105,  140,  and  245. 

3.  15,  30,  35,  70,  and  105. 

4.  174,  485,  4611,  and  5141. 

5.  30ax%  225a'xf,  and  75  aV^. 

6.  Ux"  ^  6x,  12x^  -  Sx,  Siiid  16x^  -\-  2x, 

7.  a^  -  8  b%  W-  a%  and  a%  +  4  «6^  +  2  a%''. 

8.  cc*  +  xSj  +  xhf,  4x^  -  4£cV,  and  3ic^  -  S^/l 

9.  a:;^  —  2  ic^  —  5  cc  +  6,  4  —  £c^,  and  a  —  «aj^. 

10.  a^  +  4  a^  +  16,  a"  -  4,  a«  +  8,  and  a^  -  8. 

11.  £c^  —  2  a^a?  +  ax^,  2  a^  -\-  S  a^x  -{-  ax"^,  and  4  a*x'^  —  a^cc*. 

12.  m*  -  3  mV  +  9  ti*,  m^  +  3  mn^  +  3  m^  and  3n^-^  nrn^ 
8  mTi^. 

13.  e^^  +  e-2^  -  2,  e^x  _  ^-2^,  and  e^x  _  3  _^  2  e-^^. 

14.  cc^  -  2  a;^  -  2^  -  3,  a;^  -  27,  and  x'' -^  x  -12. 


CHAPTER  III 

FRACTIONS 

28.  Operations  on  fractions.  The  four  fundamental  operations 
on  fractions  and  the  reduction  of  fractions  to  lower  or  to  liigher 
terms  depend  on  the 

Principle.  The  numerator  and  the  denominator  of  a  frac- 
tion may  he  mtdtljiUed  hy  the  same  expression  or  divided  hxj  the 
same  expression  ivithont  changing  the  value  of  the  fraction. 

29.  Changes  of  sign  in  a  fraction.  In  the  various  operations 
on  fractions  three  signs  must  be  considered,  —  the  sign  of  the 
numerator,  the  sign  of  the  denominator,  and  the  sign  before 
the  fraction.  Since  a  fraction  is  an  indicated  quotient,  the  law 
of  signs  in  division  gives  us  the 

Principle.  In  a  fraction  the  signs  of  both  numerator  and 
denominator,  or  the  sign  of  the  numerator  and  the  sign  before  the 
fraction,  or  the  sign  of  the  denominator  and  the  sign  before  the 
fraction,  may  he  changed  without  altering  the  value  of  the  fraction. 

30.  Equivalent  fractions.  Two  fractions  are  equivalent  when 
one  can  be  obtained  from  the  other  by  multiplying  or  by  divid- 
ing both  of  its  terms  by  the  same  expression. 

Two  fractions  having  unlike  denominators  cannot  be  added, 
nor  can  one  be  subtracted  from  the  other,  until  they  have 
been  reduced  to  respectively  equivalent  fractions  having  like 
denominators. 

To  change  two  or  more  fractions  (in  their  lowest  terms)  to 
respectively  equivalent  fractions  having  the  L.C.D.,  we  have  the 

Rule,  lieivrite  the  fractions  with  their  denominators  m 
factored  foi^m. 

Find  the  L.C.M.  of  the  denominators  of  the  fractions. 

34 


FRACTIONS  31 

Multiply  the  numerator  and  the  denominator  of  each  fraction 
by  those  factors  of  this  L.C.M.  which  ai'e  not  found  in  the 
deiiominator  of  the  fraction. 

If  the  L.C.D.  is  not  easily  obtained,  any  integral  multiple  of  the 
denominators  or  the  product  of  the  denominators  may  be  used.  The 
result,  however,  will  not  be  in  its  lowest  term  unless  one  uses  the  L.C.D. 

31.  Addition  and  subtraction  of  fractions.  To  find  the  algebraic 
sum  of  two  or  more  fractions  in  their  lowest  terms,  we  have  the 

EuLE.  Beduce  the  fractions  to  respectively  equivalent  frac- 
tions havijig  the  lowest  common  denominator.  Write  in  succes- 
sion over  the  lowest  common  denominator  the  numerators  of  the 
equivalent  fractions,  inclosing  each  numerator  in  a  ptarenthesis 
pjreceded  by  the  sign  of  the  corresponding  fraction. 

Rewrite  the  fraction  just  obtained,  removing  the  parentheses 
^Vi  the  nutnerator. 

Then  combine  like  terms  in  the  numerator,  and,  if  necessary, 
reduce  the  resulting  fraction  to  its  lowest  terms. 

EXERCISES 

Reduce  to  lowest  terms  : 

18 aV  _        la^-lU^    t^  x^-  -  c^b 

(x^  -  c'f 


^-     OA   ^2^.S'  ^-     .,4     I      ^,27.2  a  JA'  ^- 


24 

aV 

3.x- 

x' 

-4 

t 

7. 

2x'^- 

128 

x^'^- 

4 

8. 

x'- 

1 

2    4^         4 6 

•'       '  (x-2y  x'-c' 

3a;2  +  2a;-21 
27x^-147x-^  ' 

x"^  —  X^  -\-  X  —  1 
x'  +  a^  +  l  x'-l 

By  the  use  of  §  29  write  in  three  other  ways  : 
11.^^  13         " 


c  a  —  2  c 

2x-Zy 


12.  -—-—'' —  U 


-2  a  Zx'-^if-xy 


80  SECUAD  COURSE  iN  ALGEBRA 

Change  to  equivalent  fractions,  writing  the  letters  in  the 
denominators  in  alphabetical  order  and  making  the  first  term 
in  each  factor  positive: 

15         -3  ^  ig  ^-y 

•   (^c-a)(b-a)  '  {y-x){z-x)(z-y) 

"•1^-5^^^.  =  ?^''^   Why. 

Change  to  respectively  equivalent  fractions  having  the  lowest 
common  denominator : 

18.  f,   h   h  on      4       ^ 

19.  /5,  /^.  ah    ah 

6a3-12    a;2-2a;  ''''•  a;^  -  So;  +  6  V  -  9 

,    ^        2a;  +  5  ""      ,  2x^    ^     ,  . 

24.  Does ■—=  equal  tt— ?   Explain. 

3a  +  5  3a 

25.  Define  cancellation.    Illustrate. 

rind  the  algebraic  sum  of : 

3a      a  —  x      Sx     1/  ^^    ^  —  ^      ^ 

26-6- 9--T-  '»»-^34-7- 

a  — 3      2c  — 5  4  m^  Sm  —  x 

o^c  a(?  '  x^  —  9m^      X  -\-  Sm 

x-S      Sx^-2      a;«-8  ic  -  3  3 


21.  ^«  +  *, 
oa 

a-2h 
Aab 

x-\ 

x 

4.x'  Ux^  Sx^  x^-9x  +  14.      x' 

X  1  2-\-x 


32.  -i^ 


. /: 


x^  —  2x      X    .ic^  —  4x-j-4 

33.  2 Kr  +  T^  •  3i.  a^  +  x  +  l--^ 

V  —  11  —  r  iC  — 

35.  ^2_  f-^;^'  +0^  +  1. 

X;    —  X  -{-1 

^^      2x'-Sd^  2x  +  a 


27i»«-64a«      9a;2-16a2 


37. 
38. 
39. 


FRACTIONS  37 

3 


(a- 

-c)(x-c)       (c 

- 

aXc- 

-X) 

2x 

-4      3fl?4-l 

x 

9- 

-x'        x-3 

9 

-6x 

^x^ 

2x'  +  12c 

Sx 

■  —  7  e 

Qx'-lScx-Bc'      4x2  +  4^cx-S5c^ 


(a  —  s)(c  —  s)   ,    CLG    ,    («-  —  m)  (c  —  m) 
s  (s  —  m)  Ttis  m  (in  —  s) 


Show  that : 

42.  c  (c  —  y)  -i-  d  —  (ij  -\-  d)  d  ^  G  =  y,  when  y  =  c  —  d. 

c  —  x  —  2,  and  d^x'^  —  2x  -\-  L 
44.  =  51,  when  e^^  =  4. 

32.  Multiplication  of  fractions.  To  find  the  product  of  two  or 
more  fractions  or  mixed  expressions  we  have  the 

Rule.  If  there  are  integral  or  mixed  expressions,  reduce  them 
to  fractional  form. 

Separate  each  numerator  and  each  denoTninator  into  its  pi'ime 
factors. 

Cancel  the  factors  (^factor  for  factor)  common  to  any  numera- 
tor and  any  denominator. 

Write  the  product  of  the  factors  remaining  in  the  numerator 
over  the  product  of  the  factors  remaining  in  the  denominator. 

33.  Division  of  fractions.   For  division  of  fractions  we  have  the 

Rule.  Reduce  all  integral  or  m^ixed  expressions  to  fractional 
form. 

Then  invert  the  divisor,  or  divisors,  and  proceed  as  in  multi- 
2)lication  of  fractions.  • 


38         SECOND  COURSE  IN  ALGEBRA 

The  reciprocal  of  a  number  is  1  divided  by  the  number. 

Thus  the  reciprocal  of  2  is  ^ ;  of  ^  is  ^ ;  and  of  S^  is  3/^. 

Therefore  the  quotient  of  one  fraction  by  another  is  the  prod- 
uct of  the  first  and  the  reciprocal  of  the  second.  Also  the  quo- 
tient of  a  fraction  by  an  integral  expression  is  the  product  of 
the  fraction  and  the  reciprocal  of  the  integral  expression. 

34.  Complex  fractions.  A  complex  fraction  is  a  fraction  con- 
taining a  fractional  expression  either  in  its  numerator  or  in 
its  denominator  or  in  both. 

To  simplify  a  complex  fraction  we  have  the 

Rule.  Reduce  both  the  numerator  and  the  denominator  to 
simple  fractions,  then  perform-  the  indicated  division. 

EXERCISES 

Perform  the  indicated  operations  : 

4a;V    210"  /_a_\    /SaV  /2cy 

9a^'l0x^c'  \6cy'\c  )'\Sa')' 

/3  cy .  /9  cvy .  /-  2  a^cy  (S  a^cy 

\2a)   '  \4:a^)   '  \    5x^    )   (10  x^' 
6.  (3c  -  ac)H-(4m)  .  (8mV)--(9m  -  a^m). 

Ax"  —  2y'    4 x^"  —  Z*^    2x'-i-i/ 
^'  (2x^-ify\2x^-^y^f        3 
8    ^Jzll-x 2     \,,-^-^^  +  l        • 

^--  1  V        x^Jryx'  +  x-\-l 

10.  In  Exercise  9  multiply  each  term  of  the  first  expression 
x^  —  i, 
by  — Then  add  the  partial  products,  and  compare  this 

o 

sum  with  the  product  obtained  in  Exercise  9.    Explain.    Name 
and  state  the  law  of  nfultiplication  here  illustrated. 


FRACTIONS 

a^-a-  90 

_    ^34.9^2  ^       4a +  6 

'      a'  -  100 

«-^  +  10a  *  2(1^x^3  ax 

/          V6a\ 

a^-a-^2      2a^-12a 
a^  —  36       '     ax-{-6x 

39 


\a^  —  3  ac  —  4  cy  \  c  -\-  a    J      ^  ^ 

\«-  a^  )\  2a-\-h<i  I       \  a  a^  ) 

*"      mx  \x       mj       \   vrx""   J  \  my  \cx  —  ax  J 

17-  Multiply ^.^4 ^.^^e  •  (^=2)"^ ^y ^^^  reciprocalof ^^ 
2ic2_^5.T  +  3    3a;2  +  3aa;-j;-a   ■ 


/c4-2o^  +  (2q^  +  1  +  2c)x  _    \ 
y        x^  -\-  X  -\-  ax  -\-  a  J 

Simplify  each  term  in  : 

2     V^/7  V6  aV  2-3     \Z>7  V6  a')  ' 

21.  Explain  why   the   divisor   is    inverted   in   division   of 
fractions. 

22.  Show  how  the  rule  for  the  division  of  fractions  is  based 
on  the  first  principle  of  this  chapter. 


40  SECOND  COURSE  IN  ALGEBRA 

Simplify : 

8  —  JV  H  ^  ^ 

23.    ,,      V  •  1 ^~~ 

^-*  28.  ^.  31.  '— 

24  it±Li_±l.  1^-4  i  +  4-^ 

73     146  ^         ^^  +  c^  /4c  +  3«Y     6  c 

25  — ^ -^ " I       o  ;~ 

40J   17i  29    f 32.  i_3^J__^ 


1 


-  +  3                                    3--  33.1+  . 

27.  i 30.- -4. 1  + 


X^ 


3+      ^ 


2  ac 

m^      mri      ^ 


5-f  Tlj:jL_i 

mn 

35.  3 n^      -,1       7  5-x 

4                                 ox  — 11 — r; — 

1 ic  ox 

«  +  7^  -2+11       5  •.•^-25 

a%''  1 


(«2  _^  ^,2^)2  40. 


1- 


1 


a^  +  ^-^  ^  ~  ~ 


£C 


1  +  2^  +  5  41. 


2d 


x  + 


xy  +  l 

Y         y{x-\-xyz-\-z) 


42.  L, .     .    '       .     .■  43. 


-t- 

1 

1-a 

iC 

a;^-a^ 

1 

a  —  ic 

yz  +  1  CL      d^  +  a;^ 

z 


FRACTIONS  41 

a^  —  //  1                            1 

44.  What  value  has  —. — —^  when  a  =  e^ and  b  =  e^ -\ —  ? 

a^  +  h-  e^                         e^ 

45.  Brouncker  (1620-1684)  proved  that  tt  (the  circumfer- 
ence of  a  circle  divided  by  its 

diameter)    is  four  times   the 

fraction  on  the  right.  1  -\ 

(a)  Rewrite    the    fraction,  2  H — 

225  2  -\ — 

continuing  it  to  — r—        ,    •  ,,   ,       49 

2    +  etc.  2  H — 

(h)  Stopping  with  2  +  V-?  ^  +  ^ 
find  the    difference   in    value 


2  +etc. 


between  four  times  the  value  of  this  fraction  and  the  approxi- 
mate value  of  IT,  3.1416. 

Note.  William  Brouncker,  one  of  the  brilliant  mathematicians  of 
his  time,  was  an  intimate  friend  of  John  Wallis  (see  "  First  Course  in 
Algebra,"  page  302).  These  two  scientists  were  among  the  pioneers 
in  the  study  of  expressions  with  a  countless  number  of  terms. 

The  complex  fraction  in  the  exercise,  if  continued  indefinitely 
according  to  the  law  which  its  form  suggests,  is  called  an  infinite 
continued  fraction.  Brouncker  was  the  first  to  study  the  properties 
of  such  expressions. 


CHAPTER  IV 

LINEAR  EQUATIONS  IN  ONE  UNKNOWN 

35.  Definitions.  An  equation  is  a  statement  of  the  equality 
between  two  equal  numbers  or  number  symbols. 

Equations  are  of  two  kinds,. —  identities  and  equations  of 
condition. 

An  arithmetical  or  an  algebraic  identity  is  an  equation  in 
which,  if  the  indicated  operations  are  performed,  the  two 
members  become  precisely  alike,  term  for  term. 

A  literal  identity  is  true  for  any  value  (numerical  or  literal) 
of  the  letters  'in  it. 

An  equation  which  is  true  only  for  certain  values  of  a  letter 
in  it,  or  for  certain  sets  of  related  values  of  two  or  more  of  its 
letters/ is^ an  equation  of  condition,  or /simply  an  equation. 

A  number  or  literal  expression  which,  being  substituted  for 
the  unknown  letter  in  an  equation,  changes  it  to  an  identity, 
is  said  to  satisfy  the  equation. 

After  the  substitution  is  made  it  is  usually  necessary  to  simplify 
the  result  before  the  identity  becomes  apparent. 

A  root  of  an  equation  is  any  number  or  number  symbol 
which  satisfies  the  equation. 

36.  Axioms.  An  axiom  is  a  statement  the  truth  of  which  is 
accepted  without  proof. 

Axiom  I.  Adding  the  same  nunnher  to  each  member  of  an 
equation  does  not  destroy  the  equality. 

Axiom  II.  Subtracting  the  same  number  from  each  member 
of  an  equation  does  not  destroy  the  equality. 

Axiom  III.  Multiplying  each  member  of  an  equation  by  the 
same  number  does  not  destroy  the  equality. 

42 


LIJ^EAR  EQUATIONS  IX  ONE  UNKNOWN  43 

Axiom  IV.  Dividing  each  member'  of  an  equation  by  the  same 
number  (not  zero')  does  not  destroy  the  equality. 

Two  or  more  equations,  even  if  of  very  different  form,  are 
equivalent  if  all  are  satisfied  by  every  value  of  the  unknown 
which  satisfies  any  one. 

Of  these  four  axioms,  or  assumptions,  we  shall  make  constant 
use.  If  the  "  same  number "  referred  to  in  each  is  expressed 
arithmetically,  the  result  is  always  an  equation  equivalent  to 
the  original  one.  Further,  if  identical  expressions  involving  the 
unknown  be  added  to  or  subtracted  from  each  member  of 
an  equation,  the  resulting  equation  is  equivalent  to  the  first. 
If,  however,  both  members  of  an  equation  be  multiplied  by  or 
divided  by  identical  expressions  containing  the  unknown,  the 
resulting  equation  may  not  be  equivalent  to  the  original  one. 
In  other  words,  under  the  condition  just  stated,  roots  may  be 
introduced  or  lost  by  the  use  of  Axiom  III  or  IV  respectively. 

EXAMPLES 

1.  L£t  x-2  =1.  (1) 

Multiplying  by  a:  -  3,    a;^  -  5  a:  +  6  =  7  a;  -  21.  (2) 

From  (2),  x^  - 12  a:  +  27  =  0.  (3) 

Whence  (a:  -  3)  (a;  -  9)  ==  0.  (4) 

Therefore  a;  =  3  or  9. 

Since  (1)  has  the  root  9  only,  and  (3)  has  the  two  roots  3  and  9, 
(3)  is  not  equivalent  to  (1),  that  is,  a  root  was  introduced  by  the 
use  of  th^  multiplication  axiom. 


2.  Let                                          a:2  -  4  =  a;  +  2. 

(1) 

Dividing  by  a;  +  2,                     a:  -  2  =  1. 

(2) 

Solving  (2),                                         x  =  3. 

(3) 

But  from  (1),                      a:^  -  a;  -  6  =  0. 

(4) 

Whence                        (x  -  3)  (ar  +  2)  =  0. 

(5) 

Therefore                                             a:  =  3  or  -  2. 

Here  (5)  shows  that  (1)  has  the  two  roots  3  and  —  2,  and  since 
(2)  has  but  one  root,  3,  it  is  evident  that  a  root  was  lost  by  the  use 
of  the  division  axiom. 


44  SECOND  COURSE  IN  ALGEBRA 

The  student  should  note  the  preceding  illustration  carefully, 
as  the  possibility  of  dividing  each  member  of  an  equation  by 
a  common  factor  involving  the  unknown  frequently  arises.  A 
very  commoii'  type  is  the  following : 

3.  Let  a,-2  -  2  ar  =  0.  (1) 

Dividing  by  a:,  a:  —  2  =  0.  (2) 

Whence  a:  =  2.  (3) 

But  (1)  has  the  root  x  =  0  also,  which  is  lost  by  dividing 
both  members  of  (1)  by  x. 

If  proper  methods  of  solution  are  applied  to  an  equation  (oi- 
to  a  false  statement  in  the  form  of  an  equation),  and  one  or 
more  values  of  the  unknown  which  are  thus  obtained  do  not 
satisfy  the  original  statement,  such  values  are  called  extraneous 
(or  extra)  roots. 

An  extraneous  root  is  a  root  of  an  equation  which  is  not 
equivalent  to  the  original  statement,  but  which  is  derived  from 
it  in  the  process  of  solution. 

37.  Principles.  The  preceding  discussion  may  be  summed 
up  thus : 

Principle  I.  If  identical  expressions  (which  may  or  may  not 
contain  the  unknowri)  he  added  to  or  subtracted  from  each  mem- 
ber of  an  equation,  the  resulting  equation  is  equivalent  to  the 
original  one. 

Principle  II.  Extraneous  roots  may  be  introduced  into  a 
solution  by  multiplying  both  m^emhers  of  an  equation  by  an 
integral  expression  containing  the  unknown. 

Principle  III.  If  both  members  of  an  equation  be  divided 
by  an  integral  expression  containing  the  unknown,  one  or  more 
roots  will  usually  be  lost  by  such  a  division. 

It  can  be  seen  from  Example  1  that  the  root  introduced  is  the 
value  of  the  unknown  obtained  by  setting  the  multiplier,  a:  —  3,  equal 
to  zero  and  solving  the  resulting  equation. 

Similarly,  it  can  be  seen  from  Example  2  that  the  root  lost  is  that 
value  of  the  unknown  obtained  by  setting  the  divisor,  a:  +  2,  equal 
to  zero  and  solving  the  resulting  equation. 


■ 


LINEAR  EQUATIONS  IN  OJ^E  UNKNOWN  45 


Sometimes  a  statement  in  the  form  of  an  equation  has  no  root ; 

yet  the  ordinary  method  of  sohition  appears  to  give  one.    For  ex- 

4  a-  —  1      X  +  8 

ample,  consider  the  statement  —  = +  5. 

X  —  S       X  —  3 

If  one  multiplies  each  member  by  x  —  3  and  solves  as  usual,  he 

obtains  x  =  3.   This  answer  cannot  be  verified  because  x  —  3,  the 

denominator,  becomes  zero  for  x  =  3.    Here  the  multiplication  by 

x  —  S  introduced  the  value  3  for  x.    Checking  will  always  discover 

the  falsity  of  such  a  result  (see  "  First  Course  in  Algebra,"  page  161). 

For  solving  equations  in  one  unknown  which  may  or  may 
not  involve  fractions  we  have  the 

Rule.  Free  the  equation  of  anij  iiarentheses  it  may  contain 
except  such  as  inclose  factors  of  the  denominators. 

Where  polynomial  denominators  occur,  factor  them  if  possible. 

Find  the  L.C.M.  of  the  denoTninators  of  the  fractions  and 
multiply  each  fraction  and  each  integral  term  of  the  equation 
by  it,  using  cancellation  wherever  possible. 

Transpose  and  solve  as  usual. 

Reject  all  values  for  the  unknown  which  do  not  satisfy  the 
original  equation. 

Checking  the  solution  of  an  equation  is  often  called  testing, 
or  verifying,  the  result.    For  this  we  have  the 

Rule.  Substitute  the  value  of  the  unknown  obtained  from 
the  solution  in  place  of  the  letter  which  represents  the  unknown 
in  the  original  equation.  Then  simplify  the  result  until  the  two 
members  are  seen  to  be  identical. 

EXERCISES 

1.  Give  an  example  of  {a)  a  numerical  identity;  (5)  a 
literal  identity ;  (c)  a  conditional  equation ;  (d)  two  equivalent 
equations ;  (e)  a  statement  in  the  form  of  an  equation,  but 
which  has  no  root. 

2.  Define  and  illustrate  transposition.  («)  On  what  axiom 
does  transposition  depend  ?  (/>)  If  one  equation  is  obtained  from 
another  by  transposition,  are  the  two  equivalent  ?    Explain. 


46  SECOND  COURSE  IN  ALCJEBRA 

3.  What  extraneous  roots,  if  any,  are  introduced  if  both 
members  of  the  following  equations  are  multiplied  by  the 
expression  on  the  right? 

(a)  x-\-S  =  7  x-^4: 

(b)  x-{-5  =  0  x-\-5 

(c)  X  +  c  =  0  *  X  —  c     ^ 
{d)  X  -\-  a  =  ^  X 

(e)  x  =  5  4 

(f)  x-l=0  x^  +  Sx-\-2 

4.  What  roots,  if  any,  are  lost  by  dividing  both  members  of 
the  following  equations  by  the  expression  on  the  right  ? 

(a)  x''-4t  =  0  x-2 

(b)  x^-4:X-{-3  =  x-S  x-S 

(c)  x^-^x  —  12  =  x-\-4:  X  +  4: 

(d)  x'^  —  2x  =  4:X    .  X 

(e)  a?^  -  16  =  a;2  -  4  x^  -  4: 
.    (/)  (x  -  ay  =  (x-ay  x-a 

Solve  the  following,  for  the  unknowns  involved,  considering 
a,  b,  c,  and  d  as  known  numbers : 

3        6~4"  y      2'  714-6      4' 

5x      a       x  „3         3         7  ,^m  —  2      17 

21       6      14  2x      20      5x  m-3      18 

1,    2^  +  5      3(2^+_l)_oi 
^^-  1^  2^         "^*- 

5x-7  5/4-.t\      15a;~22 

6  2\   10   /~         6 

l_8a;  2(1- 6a;)  ^1- 24a; 

5  24a;-3    ""       15       ' 


13. 


^^    3a;  — 9       ^      3a;  — 5  _    a;   ,  x -{- m       . 

14.  ^ r  —  J  =  ^ r—  •  16.  -  H ^^ —  =  1. 

3a;  —  5  8  —  3a;  a  b 

15.1  +  1      1  =  1.  ,,.|^_^^2^ 
a      b       c      X  2a  c 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN  47 

18. \ =  — -  -  5. 

a  0  a         b 

19.         ^        +_J_^ -3(^+3). 
b(a  —  X)       a  —  X  2ab 

z  —  2      z  —  A 

^    •*^-  2,2/  +  l      22/-1      1-4/ 

3^>  +  9a;  _   g  — 2a;    _ 
9a  +  6a;      2x4-3a~    '' 

8  2-6x  27 

24. 


X  — 7      ic^-Gic  — 7      x  +  1 

25    _! 3      ^    2(^-14) 

■x  +  4      £c  —  5      cc^  —  03  —  20 

X  5  bx 

26. 


2(6^  +  ^*)       ^»-ct      U'-a"" 

»¥-s('?-'«')=-"(i-f,> 

28.  82.4 -13ic  =  32a^- 52.6. 

29.  .5704  -  .20  a;  =  19.6512  -  .016  a;. 

30.  .01  (2  X  +  .205)  -  .0125  (1.5  x  -  ;5)  =  .01955. 
2x-5.5_  7 a;^  + 3.5 a; -84.4375 

^^-  9         ~  ^  -  ^-^^  9  a;  - 11.75 

■  a;  —  .33<x      2x       ^^         4<x 
--.22« 

4a  3  5a  4a 

33. 


3  2/  -  .12  a      2       6  ?/  -  .24  a      3/  -  .04  a 


„,     6'2       1       3c -3a       1   ,    a^ 

34.  — +  -  = h-H 

ax       c  X  a      ex 


48  SECOND  COURSE  IN  ALGEBRA 

PROBLEMS 

1.  At  what  time  between  4  and  5  o'clock  will  the  hands  of 
a  clock  be  together  ? 

Solution :  First,  the  minute  hand  moves  twelve  times  as  fast  as 
the  liour  liand.  Second,  at  4  o'clock  the  hour  hand  is  20  minute 
spaces  ahead  of  the  minute  hand.  Now  let  x  equal  the  number  of 
minute  spaces  that  the  minute  hand  travels  from  its  position  at 
4  o'clock  until  it  overtakes  the  hour  hand.  Obviously  the  hour  hand 
must  travel  a:  —  20  spaces  before  it  is  overtaken  by  the  minute  hand. 

Therefore  x  =  {x  -  20)  12. 

Whence  x  =  21^\. 

Hence  the  hands  are  together  at  21^^^  minutes  after  4  o'clock. 

2.  At  what  time  between  7  and  8  o'clock  are  the  hands  of  a 
clock  together  ? 

3.  At  what  time  between  2  and  3  o'clock  are  the  hands  of  a 
clock  in  a  straight  line  ? 

4.  At  what  time  between  6  and  7  o'clock  is  the  minute  hand 
(a)  10  minute  spaces  ahead  of  the  hour  hand  ?  (b)  10  minute 
spaces  behind  the  hour  hand  ? 

5.  At  what  times  between  5  and  6  o'clock  are  the  hands  of 
a  clock  at  right  angles  ? 

6.  If  the  earth  is  between  a  planet  and  the  sun  and  in  a  line 
with  them,  the  planet  is  said  to  be  in  opposition.  The  earth 
and  Mars  revolve  about  the  sun  in  (approximately)  365  days 
and  687  days  respectively.  Mars  was  in  opposition  Septem- 
ber 24,  1909.  What  is  the  approximate  date  of  the  next 
opposition  ? 

Solution :  For  the  sake  of  simplicity  we  will  suppose  in  this  and 
in  similar  problems  that  the  planets  move  in  the  same  plane  and  in 
circular  paths,  of  which  the  sun  is  the  center. 

Let  X  =  the  required  number  of  days. 

Now  in  X  days  the  earth  will  make revolutions  about  the  sun. 

X    ^^^ 
And  in  x  days  Mars  will  make  — —  revolutions  about  the  sun. 

687 
But  to  be  in  opposition  the  earth  must  in  x  days  go  round  the  sun 
once  more  than  Mars  does. 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN  49 

Therefore  — —  = \- 1. 

365      687 

Clearing,  687  x  =  Mo  x  +  250755. 

Whence  a:  =  779  + . 

Therefore  the  required  date  is  November  11,  1911. 

7.  If  a  planet  is  between  the  earth  and  the  sun  and  in  a 
line  with  them,  it  is  said  to  be  in  conjunction.  Venus  was  in 
(superior)  conjunction  April  28, 1909.  If  Venus  revolves  about 
the  sun  once  in  225  days,  find  the  approximate  date  of  the 
next  conjunction.    ^ 

8.  Jupiter  revolves  about  the  sun  once  in  4332  days.  On 
February  28,  1909,  the  planet  was  in  opposition.  Find  the 
approximate  date  of  the  next  opposition. 

9.  Saturn  revolves  about  the  sun  once  in  10,759  days.  It 
was  in  opposition  April  3,  1909.  Find  the  approximate  date 
of  the  next  opposition. 

10.  Two  men  travel  in  the  same  direction  around  an  island, 
one  making  the  circuit  every  2^-  houi's  and  the  other  every  3 
hours.  If  they  start  together,  after  how  many  hours  will  they 
be  together  again  ? 

11.  Three  automobiles  travel  in  the  same  direction  around 
a  circular  road.  They  make  the  circuit  in  2|  hours,  3^  hours, 
and  4f  hours  respectively.  If  they  start  at  the  same  time, 
after  how  many  hours  are  the  three  together  again  ? 

12.  Is  the  answer  to  Exercise  10  an  integral  multiple  of  2^ 
and  3  ?    Is  it  the  least  integral  multiple  ? 

13.  Is  the  answer  to  Exercise  11  an  integral  multiple  of  2|, 
3^,  and  4f  ?    Is  it  the  least  integral  multiple  ? 

14.  Reduce  2|,  3^,  and  4|-  to  improper  fractions  and  divide 
the  L.C.M.  of  the  numerators  by  the  G.C.D.  of  the  denomi- 
nators.   Compare  the  result  with  the  answer  to  Exercise  11. 

15.  The  method  of  finding  the  L.C.M.  of  two  or  more  frac- 
tions or  mixed  numbers  is  hinted  at  in  Exercise  14.  State  a 
rule  therefor.    Find  by  the  rule  the  L.C.M.  of  1^,  2^,  and  3i. 


50  SECOND  COUKSE   IN  AIJiEBRA 

o  c 

16.  Find  by  the  same  rule  the  L.C.M.  («)  of  j  and  - ;  {h)  of 


a     c         -   e 
i  and 


b    d'^      f 

17.  How  many  ounces  of  alloy  must  be  added  to  56  ounces 
of  silver  to  make  a  composition  70%  silver  ? 

18.  Gun  metal  of  a  certain  grade  is  composed  of  16%  tin 
and  84%  copper.  How  much  tin  must  be  added  to  410  pounds 
of  this  gun  metal  to  make  a  composition  18%  tin  ? 

Hint.   Since  the   composition  is  16%  tin,  then  ^^/^  •  410  =  the 
number  of  pounds  of  tin  in  the  first  composition. 
Let  X  =  the  number  of  pounds  of  tin  to  be  added. 

Then  — — — \-  x  =  the  number  of  pounds  of  tin  in  the  second 

composition,  and  410  -\-  x  =  the  number  of  pounds  of  both  metals  in 
the  second  composition. 

16.10  , 

Therefore  J£^____  ^^  il    _^ 

410  +  a;        lOO'       ■ 

19.  A  30-gallon  mixture  of  milk  and  water  tests  16%  cream. ) 
How  many  gallons  of  water  has  been  added  if  the  milk  is 
known  to  test  20%  cream?     ■ 

20.  How  many  gallons  of  alcohol  90%  pure  must  be  mixed 
with  10  gallons  of  alcohol  95%  pure  so  as  to  make  a  mixtui-e 
92%  pure  ? 

21.  The  diameter  of  the  earth  is  2§  times  that  of  the  moon, 
and  the  difference  of  the  two  diameters  is  5760  miles.  Find 
each  diameter  in  miles. 

22.  The  diameter  of  the  sun  is  3220  miles  greater  than  109 
times  the  diameter  of  the  earth,  and  the  suni  of  the  two  diameters 
is  874,420  miles.    Find  each  diameter  in  miles. 

23.  The  distance  of  the  earth  from  the  sun  is  387^  times 
the  earth's  distance  from  the  moon.  Light  traveling  186,000 
miles  per  second  would  require  8  minutes  18|f  seconds  longer 
to  go  from  the  earth  to  the  sun  than  from  the  earth  to  the 
moon.    Find  each  distance  in  miles. 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN  51 

24.  The  mean  distance  between  Mars  and  the  earth  when 
they  are  on  opposite  sides  of  the  sun  is  234,500,000  miles. 
When  the  two  planets  are  nearest  each  other  on  the  same  side 
of  the  sun,  the  mean  distance  between  them  is  48,500,000 
miles.    Find  the  distance  of  each  from  the  sun  in  miles. 

25.  The  diameter  of  Jupiter  is  10|f  times  the  diameter  of 
the  earth,  and  the  sum  of  their  diameters  is  94,320  miles.  Find 
eac^.diameter  in  miles. 

26,''  A  can  do  a  piece  of  work  in  15  days  and  B  in  25  days. 
AJter  they  have  worked  together  3  days,  how  many  days  will 
B  require  to  finish  the  work  ?  -^ 

27.  A  can  do  a  piece  of  work  in  a  days,  B  in  ^  days,  and  C 
in  a  +  ft  days.  How  many  days  will  it  take  them  all  working 
together  to  do  the  work  ? 

28.  'A  cistern  has  two  pipes.    By  one  it  can  be  filled  in  2  m 

71  4-  1 
hours  ;  by  the  other  it  can  be  emptied  in  — r —  hours.    Assume 

2  m  less  than  — r—  and  find  the  number  of  hours  required  to 

o 

fill  the  cistern  if  both  pipes  are  opened. 

29.  Discuss  Problem  28  thus  :  What  is  the  relation  between 
m  and  n  if  (a)  the  water  runs  out  more  slowly  than  it  comes 
in ;  (V)  the  water  runs  out  as  fast  as  it  comes  in ;  (c)  the  water 
runs  out  faster  than  it  comjes  in? 

30.  If  both  pipes  in  Problem  28  had  been  intake  pipes,  how 
many  hom-s  would  have  been  required  to  fill  the  cistern  one-icth 
full? 

31.  If  the  radius  of  a  circle  is  increased  7  inches,  the  area  is 
increased  440  square  inches.  Find  the  radius  of  the  first  circle 
(tt  =  -2y2-  approximately). 

FacU  from  Geometry.  The  area  of  a  circle  is  the  square  of  the 
radius  multiplied  by  7r(7r  =  3.1416  approximately).  This  is  expressed 
6y  the  formula  A  =  ttR'^. 

The  circumference  of  a  circle  equals  the  diameter  times  tt.  The 
usual  formula  is  C  =  2  ttR. 


52  SECOND  COUllSE  IN  ALGEBllA 

^  32.  Imagine  that  a  circular  hoop  one  foot  longer  than  the 
circumference  of  the  earth  is  placed  about  the  earth  so  that 
it  is  everywhere  equidistant  from  the  equator  and  lies  in  its 
plane.    How  far  from  the  equator  will  the  hoop  be  ? 

33.  Compare  the  result  of  Exercise  32  with  the  one  obtained 
when  a  similar  process  is  carried  out  on  a  flagpole  6  inches  in 
diameter,  instead  of  the  earth. 

Repeating  decimals.  If  the  process  of  reducing  a  common 
fraction  to  a  decimal  does  not  end,  the  result  is  a  repeating 
decimal.  Thus  y\  =  .272727  •  •  • .  Here,  as  often  as  7  appears 
in  the  quotient  there  is  a  remainder  of  3;  that  is,  we  return 
after  each  7  to  the  original  fraction.    Then 

f^  =  .27 A,  or  ^T  =  .2727t\,  etc. 

Now  in  .27j\  the  remainder  y\  is  really  y\  of  .01  (since  7 

3  1  3 

stands  in  hundredths'  place),  or  —  of  -tttz,  or  t^-  Therefore,  if 

x  =  — ,  the  identity  —  =  .27y\  may  be  written  ic  =  .27  +  ttjt:  • 

The  relation  just  explained  will  enable  us  to  find  the  common 
fraction  which  generates  any  given  repeating  decimal. 

Dots  are  used  to  abbreviate  the  writing  of  a  repeating  decimal. 
Thus  .35  means  .3535  •  •  -,  and  .04632  means  .04632632  •  •  •. 

"  34.  Find  the  common  fraction  which,  reduced  to  a  decimal, 
gives  .393939- ••. 

Solution  :  Let  x  =  the  required  fraction. 


Then 

^  =  •^^  +  100 

Whence 

100a:  =  39  +  x. 

Solving, 

^  =  lf  =  H- 

35.  Find  the  common  fraction  which,  expressed  decimally,  is  : 
(a)  .12.  (b)  .7.  (c)  .567.  (d)  .02.   . 

(e)  3.25.    Hint.   Note  that  3  does  not  repeat. 
(/)  12.i89.  (fj)  .714285.  (A)  .142857. 


LINEAR  EQUATIONS  IN  ONE  UNKNOWN  53 

36.  A  passenger  train  whose  rate  is  42  miles  per  hour  leaves 
a  certain  station  a  hours  and  h  minutes  after  a  freight  train. 
The  passenger  train  overtakes  the  freight  in  h  hours  and  a 
minutes.    Find  the  rate  of  the  freight  train  in  miles  per  hour. 

•^  37.  The  arms  of  a  lever  are  3  feet  and  4  feet  in  length 
respectively.  What  weight  on  the  shorter  arm  will  balance 
100  pounds  on  the  longer  ? 

^  38.  A  beam  12  feet  long  supported  at  each  end  carries  a  load 
of  3  tons  at  a  point  5  feet  from  one  end.  Find  the  load  in  tons 
(excluding  the  weight  of  the  beam  itself)  on  each  support. 

39.  The  arms  of  a  balanced  lever  are  8  feet  and  12  feet 
respectively,  the  shorter  arm  carrying  a  load  of  24  pounds. 
If  the  load  on  the  longer  arm  be  reduced  4  pounds,  how  many 
feet  from  the  fulcrum  must  an  8-pound  weight  be  placed  on 
the  longer  arm  to  restore  the  balance  ? 

40.  A  horizontal  beam  12  feet  long  of  uniform  cross  section 
is  hinged  at  one  end  and  rests  on  a  support  which  is  4  feet 
from  the  other.  The  free  end  carries  a  load  of  130  pounds. 
Excluding  the  weight  of  the  beam  itself,  what  is  the  weight 
in  pounds  on  the  support  ? 

Hint.  The  products  of  the  upward  and  downward  pressures  by 
their  respective  arms  are  equal. 

41.  A  14-foot  horizontal  beam  of  uniform  cross  section  weigh- 
ing 200  pounds  is  hinged  at  one  end  and  rests  on  a  support  at 
tha  other  end.  (a)  What  is  the  weight  in  pounds  on  the  support  ? 
{h)  If  the  support  is  moved  in  3  feet  from  the  end  of  the  beam, 

f    find  the  pressure  in  pounds  on  the  support. 

^y\J\L^.  A  16-foot  horizontal  beam  of  uniform  cross  section  weighs 

fi   300  pounds.    It  rests  on  two  supports,  one  at  one  end  and  the 

other  4  feet  from  the  other  end.   Find  the  weight  in  pounds  on 

each  support. 


n 


i     2.^ 


CHAPTER  V 

LINEAR  SYSTEMS 

38.  Graphical  solution  of  a  linear  system.  The  construction  of 
the  graph  of  a  single  linear  equation  in  two  variables  or  of  a 
linear  system  in  two  variables  depends  on  several  assumptions 
and  definitions.    It  is  agreed : 

I.  To  have  at  right  angles  to  each  other  two  lines :  A''O.Y, 
called  the  j:-axis ;  and  Y'OY,  called  the  y-axis. 

II.  To  have  a  line  of  definite  length  for  a  unit  of  distance. 
Then  the  number  2  will  correspond  to  a  distance  of  twice  the 
unit,  the  number  4^  to  a  distance  4|  times  the  unit,  etc. 

III.  That  the  distance  (measured  parallel  to  the  ic-axis)  from 
the  ?/-axis  to  any  point  in  the  surface  of  the  paper  be  the 
jT-distance  (or  abscissa)  of  the  point,  and  the  distance  (measured 
parallel  to  the  ?/-axis)  from  the  ic-axis  to  the  point  be  the 
y-distance  (or  ordinate)  of  the  point. 

IV.  That  the  x-distance  of  a  point  to  the  right  of  the  ?/-axis 
be  represented  by  a  positive  number,  and  the  £c-distance  of  a 
point  to  the  left  by  a  negative  number.;  also  the  ?/-distance  of 
a  point  above  the  a;-axis  be  represented  by  a  positive  number, 
and  the  ^/-distance  of  a  point  below  the  a?-axis  by  a  negative 
number.  Briefly,  distances  measured  from  the  axis  to  the  right 
or  upward  are  positive,  to  the  left  or  downward,  negative. 

V.  That  every  point  in  the  surface  of  the  paper  corresponds 
to  a  pair  of  numbers,  one  or  both  of  which  may  be  positive, 
negative,  integral,  or  fractional. 

VI.  That  of  a  given  pair  of  numbers  the  first  be  the  measure 
of  the  £c-distance  and  the  second  the  measure  of  the  y-distance. 
Thus  the  point  (2,  3)  is  the  point  whose  cc-distance  is  2  and 
whose  2/-distance  is  3. 

54 


LINEAR  SYSTEMS 


oo 


VII.  To  call  the  point  of  intersection  of  the  axes  the  origin. 

The  values  of  the  x-  and  the  ^/-distances  of  a  point  are  often 
called  the  coordinates  of  the  point. 

The  graph  of  a  linear  equation  in  two  variables  is  a  straight  line. 
Therefore  it  is  necessary  in  constructing  the  graph  of  such  an  equa- 
tion to  locate  only  two  points  whose  coordinates  satisfy  the  equation 
and  then  to  draw  through  the  two  points  a  straight  line.  It  is  usu- 
ally most  convenient  to  locate  the  two  points  where  the  line  cuts 
the  axes.  If  these  two  points  are  very  close  together,  however,  the 
direction  of  the  line  will  not  be  accurately  determined.  This  error 
can  be  avoided  by  selecting  two  points  at  a  greater  distance  apart. 

The  graphical  solution  of  a  linear  system  in  two  variables  consists 
in  i^lotting  the  two  equations  to  the  same  scale  and  on  the  same 
axes,  and  obtaining  from  the  graph  the  values  of  x  and  y  at  the 
point  of  intersection.  Two  straight  lines  can  intersect  in  but  one 
point.  Hence  but  one  pair  of  values  of  x  and  y  satisfies  a  system  of 
two  independent  linear  equations  in  two  variables. 

Through  the  graphical  study  of  equations  we  unite  the  subjects  of 
geometry  and  algebra,  which  have  hitherto  seemed  quite  separate,  and 
learn  to  interpret  problems  of  the  one  in  the  language  of  the  other. 

(For  further  details  see  "  First  Course  in  Algebra,"  pages  187-200.) 


EXAMPLE 

Solve  graphically  the  system  2x  —  y-{-6  =  0  and  x-{-2y-\-S  =  0. 

Solution :  Substituting 
zero  for  x  and  then  zero 
for  y  in  each  equation,  we 
obtain  for  2x  —  y-{-6  =  0, 


-3 


0 


and  f  or  a;  +  2  y  +  8  =  0, 


0 
-4 

-8 
0 

Then  constructing  the 
graph  of  each  equation  as 
indicated  in  the  adjacent 
figure,  we  obtain  for  the 
coordinates  of  the  point 
of  intersection  of  the  two 
lines,  x  =—4:  and  y  =  —  2. 


1, 

/ 

/ 

/ 

/^, 

/ 

.7 

/ 

/ 

f 

-x' 

^ 

^ 

/ 

-X- 

V 

V 

_/i 

1 

_  ..., 

0 

4,- 

7 

k 

(- 

^ 

'^,9 

/ 

<: 

^ 

-•4 

/ 

<s 

N 

^ 

' 

Oli  t^ElUNi)  C(JUKhE   IN   AJ.CiKIUiA 

EXERCISES 
Solve  graphically : 

x  +  2y=l.  •  62/  +  2a;-8  =  0. 

3ic  +  42/  =  4.  '  22/-10  =  a;. 

0^  +  27/ +  11  =  0,  a. +  2/ =  5, 

""'  y-x==2.  '•  2/  +  2  =  0. 

rr  +  2y/  =  0,  2^  +  4  =  0, 

"^          82/  +  2x=:3.  2-x  =  0. 

In  Exercise  9  graph  each  equation.  Then  add  or  subtract  the  two 
equations  and  graph  the  resulting  equation  on  the  same  axes.  Note 
the  position  of  the  third  graph  with  reference  to  the  other  two. 
Proceed  in  like  manner  with  Exercises  10  and  11. 

x  +  ij  =  4.,  x-y  =  b,  Sx-4.y  =  12, 

x-^2y  =  7.  Sx-j-2y  =  5.  4:X-}-Sy  =  -6. 

12.  In  each  of  the  last  three  exercises  will  the  values  of  the 
x-  and  y-coordinates  of  the  point  of  intersection  of  the  two 
lines,  as  obtained  from  the  graph,  verify  in  the  third  equation 
obtained  by  adding  the  two  given  equations  ?    Why  ? 

13.  Graph  the  equation  x  —  2y=2.  Then  multiply  both  mem- 
bers by  2  or  3  and  graph  the  resulting  equation.  Compare  the 
two  graphs.  Then  try  —  2  or  —  3  as  a  multiplier  and  graph 
the  resulting  equation.  Compare  the  three  graphs.  What  con- 
clusion seems  warranted  ? 

14.  What  are  the  coordinates  of  the  origin  ? 

15.  Is  a  graphical  solution  of  a  linear  system  in  two  varia- 
bles ever  impossible  ?   Explain. 

16.  In  the  example  on  page  55  could  different  scales  have 
been  used  on  the  two  axes  ?    Could  the  two  lines  have  been 

•  plotted  to  different  scales  ?    Explain. 

17.  What  is  the  form  of  the  equation  of  a  line  parallel  (a) 
to  the  ic-axis  ?  (b)  to  the  ?/-axis  ? 


LINEAR  SYSTEMS  57 

18.  What  is  the  form  of  the  equation  of  a  line  through  the 
origin  ? 

19.  Give  an  example  of  a  system  in  two  unknowns  which 
has  (ci)  no  graphical  solution ;  (h)  an  infinite  number  of  sets 
of  roots. 

T^  20.  The  boiling  point  of  water  on  a  Centigrade  thermometer 
is  marked  100°,  and  on  a  Fahrenheit  212°.  The  freezing  point 
on  the  Centigrade  is  zero  and  on  the  Fahrenheit  is  32°.  Conse- 
quently a  degree  on  one  is  not  equal  to  a  degree  on  the  other, 
nor  does  a  temperature  of  60°  Fahrenheit  mean  60°  Centigrade. 
Show  that  the  correct  relation  is  expressed  by  the  equation 
C.  =  I  (F-  —  ^2),  where  C.  represents  degrees  Centigrade  and  F. 
degrees  Fahrenheit.  Construct  a  graph  of  this  equation.  Can 
you,  by  means  of  this  graph,  express  a  Centigrade  reading  in 
degrees  Fahrenheit,  and  vice  versa  ? 

21.  By  means  of  the  graph  drawn  in  Exercise  20  express 
the  following  Centigrade  readings  in  Fahrenheit  readings,  and 
vice  versa :  {a)  60°  C. ;  {h)  150°  F. ;  (c)  -  20°  C. ;  (d)  -  30°  F. 

fH    22.  What  reading  means  the  same  temperature  on  both  scales  ? 

y^  23.  A  boy  starts  at  the  southwest  corner  of  a  field  and  walks 
20  rods,  keeping  twice  as  far  from  the  south  fence  as  from  the 
west  fence.  He  then  walks  east  until  he  is  three  times  as  far 
from  the  west  fence  as  he  is  from  the  south  fence.  Lastly  he 
walks  north  until  he  is  as  far  from  one  fence  as  he  is  from  the 
other.  Construct  a  graph  of  his  path.  Find  (by  measurement) 
the  length  of  each  portion  of  it  and  his  distance  from  the 
starting  point. 

39.  Elimination.  The  process  of  deriving  from  a  system  of 
n  equations  a  system  oi  n—l.  equations,  containing  one  variable 
less  than  the  original  system,  is  called  elimination. 

If  one  equation  of  a  system  can  be  obtained  from  one  or 
more  of  the  other  equations  of  the  system  by  the  direct  ap- 
plication of  one  or  more  of  the  axioms,  it  is  called  a  derived 
equation ;  if  it  cannot  be  so  obtained,  it  is  called  independent. 


68  SECOND  COURSE  IN  ALGEBRA 

Only  two  methods  of  elimination  will  be  considered,  —  that 
of  addition  or  subtraction,  and  that  of  substitution. 

40.  Solution  by  addition  or  subtraction.  The  method  of  solving 
a  system  of  two  linear  equations  by  addition  or  subtraction  is 
illustrated  in  the 

EXAMPLE 

^  .      ^,          ^       / 130^ +  32/ =  14,  (1) 

Solve  the  system  I  ^^_^^^  22.  (2) 
Solution :  Eliminate  y  first  as  follows  : 

(1)  •  2,                                    26  a;  +  6  ?/  =  28.  (3) 

(2)  •  3,  21a:-6y  =  66.  (4) 
(3) +  (4),  47  z  =94.  .(5) 
(5) -^47,  X  =2.  (6) 
Substituting  2  for  x  in  (2),  14  -  2  y  =  22.  (7) 
Solving  (7),                                        ?/=-4. 

Check :  Substituting  2  for  x  and  —  4  for  y  in  (1)  and  (2)  gives 
r26-12  =  14,  orl4  =14. 
\14+    8  =  22,  or  22  =  22. 

The  method  of  the  preceding  solution  is  stated  in  the 

Rule.  If  necessary,  multiply  each  member  of  the  first  equa- 
tion by  a  number  and  each  member  of  the  second  equation  by 
another  number,  such  that  the  coefft,cients  of  the  same  variable  in 
both  the  resulting  equations  will  be  numerically  equal. 

If  these  coefficients  have  like  sign^,  subtract  one  equation  from 
the  other  ;  if  they  have  unlike  signs,  add  and  solve  the  equation 
thus  obtained. 

Substitute  the  value  just  found,  in  the  simplest  of  the  preced- 
ing equations  which  contains  both  variables,  and  solve  for  the 
other  variable. 

Check.  Substitute  for  each  variable  in  the  original  equations 
its  value  as  found  by  the  rule.  If  the  resulting  equations  are 
not  obvious  identities,  simplify  them  until  they  become  such. 

Or  check  in  the  sum  of  the  two  equations  unless  one  unknown 
vanishes  in  the  addition. 


LINEAR  SYSTEMS  59 

41.  Solution  by  substitution.  The  method  of  solving  a  system 
of  two  linear  equations  by  substitution  is  illustrated  in  the 

EXAMPLE 

Solve  the  system  I  g^^^^^^^g^  (2) 

Solution  :  From  (1),  3  a;  =  13  y  +  41.  (3) 

13  ?/  +  41 
Solving  (3)  for  x  in  terms  oi  y,  x  =  — ^^- (4) 

o  u  x-x  .•       13^  +  41  ,         .     ^^^ 

Substituting  — ^- for  x  m  (2), 

o 

8.i^^iii  +  ll2,  =  18.  (5) 

O 

(5)  .3,  8  (13  ?/  +  41)  +  33  y  =  54.  (6) 

SimpUfying,  104  ?/  +  328  +  33  y  =  .54.  (7) 

.    Collecting,  137?/ =-274.  (8) 

(8) -^137,  y=-2.  (9) 

—  20  +  41 

Substituting  —  2  for  y  in  (4),     a:  = =  5. 

o 

Check :  Substituting  5  for  x  and  —  2  |or  ?/  in  (1)  and  (2)  gives  the 
identities  15  +  26  =^  41  and  40  -  22  =  18. 

The  method  of  the  preceding  solution  is  stated  in  the 

Rule.  Solve  either  equation  for  the  value  of  one  variable  in 
terms  of  the  other. 

Substitute  this  value  for  the  variable  in  the  equation  from 
which  it  teas  not  obtained  and  solve  the  resulting  equation. 

Substitute  the  definite  value  just  found,  in  the  simplest  of  the 
preceding  equations  which  contains  both  variables,  and  solve,  thus 
obtaining  a  definite  value  for  the  other  variable. 

Check.  As  in  preceding  example. 

If  either  equation  in  a  linear  system  contains  fractions,  it  is 
usually  (see  exception  in  Exercise  22,  page  61)  best  to  clear  of 
fractions  and  then  to  apply  one  of  the  preceding  rules.  Occa- 
sionally one  can  avoid  quadratics  by  solving  without  first 
clearing  of  fractions  (see  Problem  7,  page  77). 


60  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

1.  What  is  a  constant  ?  a  variable  ? 

2.  Define  and  give  examples  involving  two  unknowns  of 
(rt)  a  linear  equation ;  (b)  a  system  of  linear  equations ;  (c)  a 
simultaneous  linear  system;  (d)  equivalent  equations;  (e)  a 
determinate  system;  (/)  an  indeterminate  equation;  (g)  an 
indeterminate  system;  (h)  an  incompatible  or  inconsistent 
system.  ^ 

3.  What  is  the  graplrrireach  case  (except  d)  in  Exercise  2  ? 

4.  What  is  the  general  form  of  a  linear  equation  in  two 
variables  ? 

5.  To  what  general  form  may  any  incompatible  linear  sys- 
tem in  two  unknowns  be  reduced  ? 

6.  What  is  the  general  form  of  a  linear  system  in  two 
unknowns  which  has  an  infinite  number  of  sets  of  roots  ? 

Solve  by  addition  or  subtraction : 

/     2x  +  5y=:8,  9t-2n  =  lS, 

x-10y  =  9.  20^  =  771  +  63. 

50-4-38  =  12?/,  llm-10  =  -187i, 

3x-}-Sy  =  0.  97n  +  12n  =  -15. 

11.  3a; -22/ =  18,  30  +  8?/  =  5x.      / 

Solve  bv  substitution :  / 

13    2(a:  +  ?/)  +  3?/  =  4,  7?/-3^-4  =  0. 

'  5  =  x-\-y.  Sm-3n  ,   ^     _      q 

,,    160^  +  7  =  15?/,  16.  2         +^^-      y^ 

14. 

*  4  ic  +  5  ?/  =  0.  4  ??^  —  1  =  3  71. 


Solve  by  either  one  of  the  preceding  methods : 

17.    3   "^^"'        '  18.    4       2~12 

5a; -3?/ =  105.  r  +  8=-2, 


LINEAR  SYSTEMS  61 

Sy  +  1      z+20^  3  4   ^ 

89"  ^5~5 


20. 

5x-^.7y  =  -l. 

25, 

.12x-.0Sy  =  2 

•f- 

m,  -\-  2n      2  m  — 

71      5 

21. 

5                 10 

2 

m-^  n      m  —  n 

n 

4               7 

2 

22. 

~x^y=-V 
2      3_      4 
a;      ?/          3 

23. 

2r  +  4s      38 
2r-s   ~~  3  ' 
5      14 

r        s 

2  7n  +  3n-2 

4 

24. 

m      n      m 

3^ 

^0.^^  +  %2- 

12 

26. 


.25  h  +  ^-^lk  _^^ 
1  20 


.8A-2.2  ■  35-5.5A: 
1  =  3. 


=  0. 


X  —  n 


k 

-^ 

3  (a 

•  +  ?/) 

1   «-2/ 

= 

0, 

28. 

f 

'    -V 

2£C 

+  2/  = 

=  7. 

5~ 

.^     : 

2 
.1   1 

y^— 

5 

A 

29. 

1 

1 

1 

—  ^y 

x 
4~ 

-22/  = 

13 

'  2  ' 

-+%: 

'J  +  "      0. 

m  -\-  5  _  m  —  5     m  — |(3m  — 2?i)— i       ^  _  n 

31.  —  77:?  '         ^:^  ; -f- -  —  0. 

n  7i  — 10  7/Z-  —  Jti  —  1  3 


Solve  for  x  and  ?/ : 

2  ex      y 

^  =  5c, 

3        c 

33.  "-      ' 

2«  +  3c-^^- 

a      Sa  _ 

34.  ^     ^     ^ 

0  a       a       1 

62 


y +  ?/  ^  9  _  ^  —  y 

35.  10  a   "    4  a 

x  =  y. 

x  y 


SECOND  COURSE  IN  ALGEBRA 

(c-\-b)y-^ax  =  l, 


36. 


X  y 


37. 


38. 


39. 


«?/  =  1  - 

■(c-^b)x. 
=  a^, 

ax  +  by 
dx  -f-  ey 

42.  Determinants  of  the  second  order.    The  arrangement  of 

4     2 

has  been  given  the  meaning  4  •  3  —  5  •  2.  ^^    • 


numbers 


5     3 


Such  an  arrangement  is  called  a  determinant. 
The  value  of  any  such  determinant  is  easily  found  since 
means  ad  —  be. 

5     6 


Accordingly, 
Similarly, 
And    3 


2     3 
6-^2 
8       3 
3         9 
-4     -5 


=  5-3-2. 6  =  15 -12  =  3. 

=  6 . 3  -  8  •  (-  2)  =  18  +  16  =  34. 


3[-  15  -  (-  36)]=  3[-  15  +  36] 
=  3  •  21  =  63. 

The  preceding  operations  can  be  reversed  and  the  difference 
(or  sum)  of  two  products  written  as  a  determinant. 

m     s 


Thus  mn  —  i^s  can  be  written 
other  ways. 

Similarly,     ab  —  k  =  ab  —  Ik  = 


n 

a  '  Jc 
1     b 


,  and  in  a  number  of 


EXERCISES 

Find  the  value  of  the  determinants : 


4     1 

1. 

3     5 

• 

6-2 

VI 

8 

1 1 

4. 

— 

-3     2 

7       8 

5. 

3 

i  6 
i     8' 

7. 

2a 

-10/> 

5  a 

?>b 

„     5 

a' 

0 

8. 

2a 

a" 

2a 

3.  3 


2       3 
4-3 


6.  4 


-3     5 


'■I 


Ze  0      I 

9d     -Ue^l 


LINEAR  SYSTEMS 


Write  as  a  determinant : 

10.  ax  —  cr.  12.  §  —  ar. 

11.  mz-Zd.  13.  hk-G. 

Find  the  value  of  the  fractions : 


14.  ah  -|-  cd. 

15.  a-h. 


16. 


1        1 
1      -1 


17. 


3  2 

4  3 


18. 


c 

h 

f 

e 

a 

h 

d 

e 

19. 


c 

{1-e) 

(7 -12  c) 

c       3 

7  c     36 

Write  as  the  quotient  of  two  determinants 


20. 


21. 


st- 

-cd 

?>cx 

-  5  r 

a  + 

10 

c 

a'  -  12 


22. 


23. 


ax 

-6 

2y  - 

-^t 

m 
3" 

-7 

24. 


25. 


7fl  —  1 


6x  —  hm 

3  tt  —  3  r 

o^      15* 

4  a 


43.  Solution  by  determinants.    For  the  system 
ax  -\-  by  =  c, 
dx-\-ey=f, 


(1) 
(2) 


ae  —  hd 


-^   and  2/  =  -"^ 


cd 


hd 


(see  Exercise  39,  page  62). 


Using  determinants,  x  = f- 


ae 


and 


2/  = 


af  —  cd 
ae  —  hd 


A 

a  c 
d    f 

a  h 
d     e 

(3) 


w 


The  determinant  expressions  for  x  and  y  in  (3)  and  (4)  can 
be  used  as  formulas  to  solve  any  pair  of  linear  equations  in 
two  unknowns.  This  method  is  particularly  useful  in  the  solu- 
tion of  linear  equations  with  literal  coefficients.  The  determi- 
nant forms  can  be  easily  remembered  and  written  down  at  once 
if  we  observe  carefully  the  following  points  : 


64  SECOND  coursp:  in  algebra 

I.  The  determinants  in  the  denominators  are  identical,  and 
each  is  formed  by  the  coefficients  of  x  and  y  as  they  stand  in 
the  original  equations  (1)  and  (2). 

II.  The  determinant  in  the  numerator  of  the  value  of  x  is 
formed  from  the  denominator  by  replacing  the  coefficients  of  x, 
a  c 

d,  by  the  constant  terms  f, 

III.  The  determinant  in  the  numnerator  of  the  value  of  y  is 
formed  from  the  denominator  by  replacing  the  coefficients  of 

b  c 

y,  e,  by  the  constant  terms  f. 

Biographical  Note.  Gottfried  Wilhelm  Leibnitz.  For  the  last  few 
hundred  years  the  study  of  the  higher  mathematics  has  been  carried  on 
almost  entirely  hy  professors  in  the  universities.  It  is  rather  exceptional 
for  a  man  not  connected  with  any  educational  institution  to  achieve  dis- 
tinction in  this  field.  Before  this  was  the  case,  however,  scholars  were 
accustomed  to  devote  themselves  to  any  or  all  branches  of  learning  which 
attracted  them,  and  many  men  of  wide  erudition  in  various  walks  of  life 
flourished  at  different  times  during  the  two  or  three  hundred  years  fol- 
lowing the  fifteenth  century. 

But  of  them  all,  the  man  who  perhaps  most  clearly  deserves  the  title 
of  universal  genius  is  Leibnitz  (1646-1716).  He  was  born  in  Leipzig, 
Germany,  and  on  account  of  the  poor  instruction  in  the  school  to  which 
he  was  sent,  he  was  obliged  to  learn  Latin  by  himself,  which  he  did  at 
the  age  of  eight.  By  the  time  he  was  twelve  he  read  Latin  with  ease,  and 
had  begun  Greek.  Not  until  the  age  of  twenty-six,  when  he  was  sent  to 
Paris  on  a  political  errand,  did  he  become  deeply  interested  in  mathe- 
matics. From  1676,  for  nearly  forty  years,  he  held  the  well-paid  posi- 
tion of  librarian  in  the  ducal  palace  of  Brunswick,  serving  under  three 
princes,  the  last  of  whom  became  George  I  of  England  in  1714.  This 
post  afforded  him  time  for  the  deep  study  of  mathematics,  philosophy, 
theology,  law,  politics,  and  languages,  in  all  of  which  he  distinguished 
Jiimself.  An  incomplete  edition  of  his  mathematical  works  has  been 
published  in  seven  volumes. 

Personally  he  was  quick  of  temper,  impatient  of  contradiction,  over- 
fond  of  money,  and  one  of  the  few  really  great  men  who  have  been 
offensively  conceited. 

It  is  in  his  writings  that  we  find  the  first  mention  of  determinants. 
He  also  discovered  the  calculus  independently  of  Sir  Isaac  Newton,  and  the 
last  years  of  both  men  were  embittered  by  a  most  unfortunate  wrangle  in 
which  the  friends  of  Newton  accused  Leibnitz  of  publishing  as  his  own, 
results  which  really  belonged  to  Newton. 


GOTTFRIED  WILHELM  LEIBNITZ 


LINi:AR  SYSTEMS 


65 


EXAMPLE 


Solve  by  determinants 


2ij-\-x=^l,  .        • 

5x  =  22/H-ll. 
Solution :  Writing  the  equations  in  the  standard  form,  we  have 


Then 


5  a.- 

-2.y 

7 

2 

11 

-2 

1 

2 

5 

_  2 

11. 


- 11  -  22 
-2-10 


36 


12 


=  3. 


In  solving  for  y  the  denominator  is  the  same ;  hence 

1      7 


y  = 


11 


11  -  35      -  24 


12 


12 


12 


-c-lt^-i. 

EXERCISES 

Solve  by  determinants  and  check  results  : 

^'  3x-2y  =  i. 

--^-6 
2      3~    ' 

2^3 

4;^=:.  37/  +  8, 

''•  5ij-\-6  =  3x. 

5x  +  42/  =  10a  +  4, 
•  x-2aij  =  0. 

7  X  4-  5?/  =  21c 

^•^-f  =  3. 

.3x-\-.02y  =  lS5, 
-  .5x4- .04  7/ =  335. 
5    4x  +  37/  =  6, 

c       2c 

X       7/       a  4-  ^^ 

X  —  7/  =  ; 

44.  Indeterminate  equations.  If  numerical  values  are  given 
to  any  two  variables  in  the  equation  7ii  +  n  +  p  =  6,  a  value 
for  the  third  variable  can  be  found,  which,  taken  with  the 
values  assigned  to  the  other  variables,  satisfies  the  equation. 

For  example,  let  m  =  1  and  7i  =  2.  Then  m  -\-  n  +  p  =  Q  becomes 
^  4  2  +  j9  =  6,  whence  p  =  3.     Obviously  jh  =  1,  n  =  2,  and  p  =  d 


00 


JSECOAD  CUUK«E  l.\   ALGEJBKA 


satisfy  the  equation.  Other  values  may  be  given  to  m  and  n  (or  m 
and  p,  or  n  and  7;),  and  the  foregoing  process  repeated,  thus  obtain- 
ing set  after  set  of  roots.    A  few  sets  of  roots  are  tabulated  here. 


m 

1 

2 

0 

1 

6 

\ 

-4 

10 

n 

0 

2 

0 

3 

0 

1 

2 

-1 

P 

3 

2 

6 

2 

0 

3 

8 

-3 

In  like  manner,  for  2  m  +  3  w  +  4/»  — 16  the   following  sets  of 
roots  may  be  obtained  : 


B 


In  tables  A  and  B  the  values  of  m  and  n  are  alike,  but  the  corre- 
sponding values  of  p  are  different  in  every  case.  Though  each  equa- 
tion has  an  infinite  (unlimited)  number  of  sets  of  roots,  no  common 
set  appears  in  the  tables.    This  suggests  three  questions : 

(1)  Have  the  two  equations  a  common  set  of  roots  ? 

(2)  If  so,  is  there  more  than  one  common  set  ? 

(3)  Is  the  number  of  such  sets  of  roots  unlimited  ? 
These  questions  are  answered  by  the  work  which  follows. 


m 

1 

2 

0 

1 

G 

\ 

-4 

10 

n 

2 

2 

0 

3 

0 

^ 

2 

-1 

P 

2 

H 

4 

1 

1 

n 

H 

-i 

01      XI,  i.         (m  -\-  n  -^  p  =  Q, 

Solve  the  system  •{  ^       ,  q      ,   a         1 « 
•^  ^^2  m -f  3  n -h  4/>  =  16. 

Solution :  First  eliminate  one  variable,  say  m,  as  follows 
(1).2,  2m  +  2n  +  2p  =  12. 

(2)  —  (3)  eliminates  m,  giving  n  -^  2p  =  i. 

Give  n  or  JO  in  (4)  any  value,  say  «  =  8,      8  +  2^^  =  4. 
Solving  (5),  p  =—2. 

Substituting  8  for  n  and  —  2  for jo  in  (1),  //i  -f-  8  —  2  =  6. 
Solving  (7),  7)1  -  0. 

Therefore  the  set  m  =  0,  n  =  8,  and  p  =—2  satisfies  (1). 
These  values  also  satisfy  (2), 
since  0  -1-  24  -  8  =  16,  or  16  =  16. 

Again,  letp  =  4  in  (4),  ?i  +  8  =  4. 


(1) 
(2) 

(3) 
(4) 
(5) 
(6) 
(7) 
(8) 


(9) 


LINEAR  SYSTEMS 


67 


VI 

2 

4 

0 

H 

n 

4 

0 

8 

I- 

2 

-1 

_  2 

P 

f)        2 

_2 

H 

5 

2 

-1 

-10 

Solving  (9),  n=-4:.        (10) 

Substituting  —  4  for  n  and  4  forp  in  (1),  m  —  4  +  4  =  6.  (11) 

Solving  (11),  m  =  6.  (12) 

Then  m  =  Q,  n  =—  i,  and p  =  4:  satisfy  (1). 
They  also  satisfy  (2),  since  12  - 12  +16  =  16. 

Repetition  of  the  foregoing  process  gives  the  four  following  sets 
of  roots  for  the  system  (1)^.(2): 


C 


Since,  by  the  method  just  explained,  table  C  may  be  completed 
and  extended  as  much  farther  as  is  desired,  we  conclude  that  the 
system  (1),  (2)  has  an  infinite  (unlimited)  number  of  sets  of  roots. 
We  must  not  infer,  however,  that  every  pair  of  linear  equations  in 
three  variables  has  either  an  infinite  number  of  sets  of  roots  or  even 
one  set.    Such  an  inference  is  easily  seen  to  be  incorrect. 

For  instance,  let  x  +  y  +  z  =  8,  (1) 

and  X  i-  y  +  z  =  24:.  (2) 

An  attempt  to  eliminate  one  variable  by  subtracting  (1)  from  (2) 
causes  all  the  variables  to  vanish  and  gives  0  =  16,  which  is  false. 
Hence  the  method  of  the  last  solution  fails. 

In  general,  however,  a  system  of  two  independent  equations  of 
the  first  degree  in  three  variables  has  an  infinite  (unlimited)  num- 
ber of  sets  of  roots. 

A  system  of  three  independent  equations  of  the  first  degree  in  three 
variables,  no  two  equations  being  incompatible,  has  one  set  of  roots 
and  only  one. 

Note.  It  is  not  a  little  remarkable  that  the  writings  of  the  first 
great  algebraist,  Diophantos  of  Alexandria  (about  300  a.d.),  are 
devoted  almost  entirely  to  the  solution  of  indeterminate  equations ; 
that  is,  to  finding  the  sets  of  related  values  which  satisfy  an  equa- 
tion in  two  variables,  or  perhaps  two  equations  in  three  variables. 
We  know  practically  nothing  of  Diophantos  himself,  excepting  the 
information  contained  in  his  epitaph,  which  reads  as  follows :  "Dio- 
phantos passed  one  sixth  of  his  life  in  childhood,  one  twelfth  in 
youth,  one  seventh  more  as  a  bachelor ;  five  years  after  his  marriage 
a  son  was  born  who  died  four  years  before  his  father,  at  half  his 


68  SECOND  COURSE  IN  ALOEBRA 

father's  age."  From  this  statement  the  reader  was  supposed  to  be 
able  to  find  at  what  age  Diophantos  died.  As  a  mathematician  Dio- 
phantos  stood  alone,  without  any  prominent  forerunner,  or  disciple, 
so  far  as  we  know.  His  solutions  of  the  indeterminate  equations 
were  exceedingly  skillful,  but  the  methods  which  he  used  were  so 
obscure  that  his  work  had  comparatively  little  influence  upon  that 
of  later  times.  Ae  ^  #^w 

45.  Determinate  systems.    The  method  of  obtaining  the  set 
of  roots  of  a  determinate  system  is  illustrated  in  the  following 

EXAMPLE 

(m  +  &n-^p  =  2Z,  (1) 

Solve  the  system  -j  3  m  —  8  ?i  _+  4p  =  —  1,  (2) 

[7m-10?i  + 10^  =  0.  (3) 
Solution  :  Eliminate  one  variable,  say  p,  between  (1)  and  (2)  thus  : 

(1) .  4,                                                   4  m  +  24  n  -  20  p  =  92.  (4) 

(2).  5,                    '                             15m-40n  +  20;?  =-5.  (5) 

(4)  +  (5),                                           19  m- 16  n              =  87.  (6) 
Now  eliminate  p  between  (1)  and  (3)  as  follows : 

(l)-2,                                                   2  wi  +  12  n  -  lOjo  =  46.  (7) 

(3).l,                                           '       7m-10n  +  10;?  =  0.  (8) 

(7) +  (8),                                             9m+    2n              =46.  (9) 
The  equations  (6)  and  (9)  contain  the  same  two  variables  m  and  ti. 

(6)-l,                                                             19w-16n  =  87.  (10) 

(9) -8,                                                             72  m  +  16  n  =  368.  (11) 

(10)  4- (11),                                                   91m              =45.5.  (12) 

(12) -^91,                                                                       m  =  5.  (13) 

Substituting  5  for  m  in  (6),                           95  -  16  n  =  87.  (14) 
Solving  (14),                                                                   ^  —  -k' 

Substituting  ^  for  n  and  5  for  m  in  (1),  5  +  3  —  5/>  =  23.  (15) 
Solving  (15),                                                                   jw  =  -  3. 
Check:  Substituting  6  for  m,  ^  for  n,  and  —  3  for  ;?  in  (1),  (2), 

^^^^  ^^^'  5  +  3  +  15  =  23,  or  23  =  23. 

15-4-12=-l,  or  -1  =  -1. 
35  -  5  -  30  =  0,  or  0  =  0. 
Or  we  may  check  thus : 

(1)  +  (2)  +  (3)  gives  llm-12n  +  9p  =  22. 

Substituting  for  m,  n,  and  p,  55  -  6  -  27  =  22,  or  22  =  22. 


LINEAR  SYSTEMS  69 

This  last  check  fails  if  any  of  the  variables  vanish  when  the  three 
equations  are  added. 

For  the  solution  of  a  simultaneous  system  of  linear  equations 
in  three  variables  we  have  the 

Rule.    Decide  from  an  inspection  of  the  coefficients  which 
variable  is  inost  easily  eliminated. 
J    Using  any  two  equations,  eliminate  that  variable. 

With  one  of  the  equations  just  used,  and  the  third  equation, 
again  eliTninate  the  same  variable. 

The  last  two  ojyerations  give  two  equations  in  the  same  two 
variables.    Solve  these  equations  by  the  rule,  i^age  58. 

Substitute  the  tivo  values  found  in  the  simplest  of  the  origin 
nal  equations  and  solve  for  the  third  variable. 

Check.  Substitute  the  values  found  in  each  of  the  original 
equations  and  simplify  results. 

Or  check  in  the  sum  of  the  three  original  equations  (unless 
one  variable  vanishes  in  addition). 

Four  or  more  independent  equations  in  three  variables  have 
no  common  set  of  roots. 

In  general,  a  system  .of  n  +  1  independent  linear  equations  in  n  varia- 
bles has  no  set  of  roots :  a  system,  of  n  independent  linear  equations  in  n 
variables,  no  ttoo  of  tchich  are  incompatible,  has  one  set  of  roots ;  and 
a  system  ofn  —  1  independent  linear  equations  in  n  variables,  no  ttoo  of 
ivhich  are  incompatible,  has  an  infinite  number  of  sets  of  roots. 

The  usual  proof  of  the  preceding  theorem  affords  a  beautiful 
application  of  determinants.  Though  not  extremely  difficult,  it 
requires  greater  familiarity  with  determinants  than  the  student 
will  acquire  from  this  text. 

A  system  of  four  independent  equations  in  four  variables 
may  be  solved  as  follows  : 

Use  the  first  dnd  second  equation,  then  the  first  and  third, 
and  lastly  the  first  and  fourth,  and  eliminate  the  same  variable 
each  time.  This  gives  a  system  of  three  equations  in  the  same 
three  variables,  which  can  be  solved  by  the  rule  given  above. 


70  SECOND  COURSE  IN  ALCiEBllA 

EXERCISES 

1.  Find  five  sets  of  roots  iov  x  —  2 1/  -\-  z  =  6. 

2.  Find  five  sets  of  roots  for  7n  —  n  —  2p  =  S. 

3.  Fill  out  the  blanks  in  table  C,  page  67. 

4.  Find  three  sets  of  roots  for  the  system 

m  +  n  —  p  =  S, 

3  m  —  2  ?i  +  4  j9  =  6. 

Solve  the  following  systiems  : 

2x-\'Si/  =  -U-4:Z,    .        .4r  +  .3s-8^^4,       v 

5.  x-y-\-Sz  =  0,  7.  .5r  +  ^4-.8s  =  1.2, 
5x  +  z  =  U-2y.  2.6t-\-.S-r  =  -{-.5s. 

X  -f-  2y  +  3.^  =  14,  .25ar  +  .05 //  =  -  1  +  .10^, 

6.  4cc-52/  +  6;s  =  12,  8.  .50 a;  -  .30 7/ =  0,       . 
x  +  lhy-\-^z  =  58.  .057/  +  .04.i;  =  3. 

In  Exercises  9,  10,  and  11  consider  a,  h,  c  as  known  numbers : 

I  _|_  i£  _  _L  =  6  3  //?.  +  2p  =  6  a  -  2  71, 

a,     2a      ^a        '  10.  7/1  -  5n  +  6^  =  2  r/ -  11 /;, 

9.  77'  +  4^=6.s  6w^-8p  =  12a  +  8/>. 

9      6 

h  +  2k-l  =  ^h  +  c, 

h  _  3  a  +  h       k_2c^ 
2~"       6       "^3        6  ' 

In  Exercises  12-14  solve  for  x,  y,  and  z : 

a^.a^y.a^  =  a-\  a       h       r       ^ 

12.  c2^-c-2'-c^  +  i  =  6'-i^  i"^^"^,^""    ' 

14     ^_^    .    ?_^==9 

JyC.JjU  +  2^112^  •"*•     X  7/  ;^ 

13.  6-^  •  C*  +  4  =  6-16,  ^  _  2C  _  ?i 

dv-d''  +  ^  =  (]}\  X        '-  ~  y  ' 


LINEAR  SYSTEMS 


n 


15 


7'  -}-  s  -\-  t  -{-  u  =  6, 

5r -\- 9s -[-4.ti  =  SI  -{-6t, 
r-^9t  —  7u=—54.  —  5s. 

(In  Exercise  18  solve  for  x  only.) 
17.  5x-\-4:y-10z  =  3, 


16. 


18. 


34.z-7x-lSy  =  31. 


46. 


4  A  —  A  +  m  =  0,^ 

16  7*  +  5  /c  -  X  =  4. 


ax  -\-hy  -\-  cz  —  p, 
dx  +  ey  +  fz  =  q^ 
(jx  +  A?/  +  iz  =  r. 

The    arrangement   of 


numbers. 


Peterminants  of   the  third  order. 

lias  been  given  the  meaning  5  •  2  •  3  +  7  •  8 


"^^H 


6 


1Q8 


4(\ 


4-  9 , 1 . 4  _  6  -'2  .  9  -  1  •  8-  5  -  3  •  7  •  4,  which  equals  30  +  336 
+  36  -  108  -  40  -  84  =  170. 

Such  an  arrangement  is  called  a  determinant  of  the  third 
order  because  it  has  three  rows  (horizontal  lines  of  numbers) 
and  three  columns  (vertical  lines  of  numbers).  Each  of  the 
nine  numbers  in  the  determinant  is  called  an  element. 

Every  determinant  of  the  third  order  is  equal  to  a  poly- 
nomial of  six  terms. 
Each  of  the  six  terms 
is  the  product  of  three 
elements  so  chosen 
that  one  element,  and 
only  one,  is  taken  from 
each  row,  and  one  ele- 
ment, and  only  one,  is 
taken  from  each  col- 
umn. If  each  element  is  positive,  three  terms  of  the  polynomial 
are  positive  and  three  are  negative.  In  connection  with  the  pre- 
ceding explanation  the  student  should  study  carefully  the  above 
diagram,  in  which  each  continuous  line  connects  three  numbers 
wdiose  product  gives  a  positive  term,  and  each  dotted  line  con- 
nects three  numbers  wh^se  product  gives  a  negative  term. 


i'2  SECOND  COURSE  IN  ALGEBRA 

It  follows,  then,  that  the  preceding  determinant  is  equal  to 
the  polynomial 

5. 2. 3  +  7-8.  6  +  91. 4-6-2. 9-1- 8. 5-3. 7-4  = 

30         +336      +36         -108       -40         -84         =170. 

When  finding  the  value  of  each  product  in  a  determinant, 
the  sign  of  every  negative  element  must  be  taken  into  account 
along  with  the  foregoing  explanations. 


Find  the  value  of : 


EXERCISES 


2. 


1>  I  % 

1  0 

2  4 

3  1 

0 
5 

1 

• 

v.^9. 

X 

2 
3 

y   1 

2     1  . 
4     1 

12    3 

Ill 

a 

h     c 

4     5     6 

,6. 

a     1     a 

. 

10. 

d 

e    /. 

7     8     9 

—  a     5     6 

9 

h     i 

1-2     3 

2-13 

X 

y    1 

3         2     1 

.  V  7. 

— 

-3       12 

.      11. 

x^ 

2/i    1- 

1         11 

4      5     1 

X, 

y.   1 

a     h      c 

a     2     1 

5 

1   h 

12     3 

8. 

h     3     8 

12. 

c 

c 

0     0 

a     h      c 

G 

4     9 

8 

6     a 

47.  General  linear  system^ 

ax 

dx 

gx 

pei  +  qhc 


y 


. .  del  +  dhc  +  gfh  —  ceg  —fha  —  idb 
aqi  +  drc  +  gfp  —  cqg  —  fra  —  idp 
aei  +  dhc  +  gfb  —  ceg  —  fha  —  idb 
aer  +  dhp  +  ^(/Z*  —peg  —  qha  —  rdb 


aei  +  dhc  +  gfb 
(See  Exercise  18,  page  71.) 


ceg  —fha  —  idb 


(4) 
(5) 
(6) 


LINEAR  SYSTEMS 


73 


In  the  fractions  in  (4),  (5),  and  (6)  observe  the  following 
points : 

1.  The  three  denominators  are  identical. 

2.  Each  numerator  and  each  denominator  contains  six  terms, 
three  positive  and  three  negative.    ' 

3.  Each  term  is  the  product  of  three  letters,  one  of  these 
letters,  and  only  one,  being  taken  from  each  equation. 

4.  Each  term  in  the  rumierator  differs  from  the  term  just 
below  it  in  the  denominator  ^by  one  letter,  and  only  one. 

The  preceding  statements  will  help  to  make  clear  the  reason 

for  what  now  follows. 

Let  us  write  the  coefficients  of  x,  y,  and  s  as  a  determinant 

in  the  order  in  which  they  occur  in  (1),  (2),  and  (3),  and  then 

expand.    We  thus  obtain 

a     h      c 

d     e     f 

g     h     i 

\  ^  .  .       . 

But  (Z))  —  aei  -\-  dhc  -f-  gf^  —  c^g  —  fha  —  idb,  which  is  the 

denominator  of  the  fractions  in  (4),  (5),  and  (6). 


<p 


If  in  (Z))  we  now  replace  the  coefficients  of  x,  d,  by  the  con- 

P  g 

stant  terms  g,  and  expand,  we  obtain  a  determinant  equivalent 

r 
to  the  miTYierator  of  the  fraction  (4)  luJiose  value  is  x  ;  for 
p    b     c 

q     e    f  =  pel  +  q^ic  +  t]fb  —  cer  —flip  —  iqh. 
r     h    i  ij 

Again,  if  in  (Z>)  we  replace  the  coefficients  of  y,  e,  by  the 

constant  terms  g,  we  obtain  a  determinant  equivalent  to  the 

r 
numerator  of  the  fraction  (5)  whose  value  i^'y.  q 

Lastly,  if  in  (7))  we  replace  the  coefficients  of  z,  /,  by  the 

P  i 

^  constant  terms  g,  we  obtain  a  determinant  equivalent  to  the 


74 


iSECUAi)   COUK8E  lA'  ALGEBRA 


numerator  of  the  fraction  (6)  whose  value  is  z.    (The  student 
should  perform  the  work  outlined  in  the  last  two  sentences.) 
Therefore  we  may  write  the  values  of  cc,  y,  and  z  for  the 
given  system  in  determinant  form  as  follows : 


p 

h 

c 

Q 

e 

f 

r 

h 

I 

a 

h 

c 

d 

e 

f 

U 

h 

I 

U)    y  = 


a 

P 

c 

d 

Q 

f 

9 

r 

I 

a, 

h 

c 

d 

e 

f 

9 

h 

i 

(8) 


d 

h 

P 

d 

e 

Q 

9 

h 

r 

a 

h 

c 

d 

e 

f 

9 

h 

I 

(9) 


The  fractions  (4),  (5),  and  (6)  are  general  results,  and  can 
be  used  as  formulas  to  solve  any  three  simultaneous  equations 
in  three  variables,  but  the  equivalent  forms  (7),  (8),  and  (9) 
are  far  more  easily  remembered.  These  can  be  written  down 
at  once  for  any  system  of  three  equations  in  three  variables, 
since 

I.  The  determinants  in  the  denominators  are  identical,  and 
each  is  formed  hy  the  coefficients  of  x,  y,  and  z,  as  they  stand 
in  the  original  equations. 

II.  Each  determinant  in  the  numerator  is  formed  from  the 
denom^inator  hy  2^uttlng  the  colum^n  of  constant  terms  (as  they 
stand  in  the  original  equation^  in  place  of  the  column  of  the 
coefficients  of  the  variable  whose  value  is  sought. 

The  method  of  solution  by  determinants  of  a  system  of 
equations  in  three  variables  is  illustrated  in  the  following 


EXAMPLES 

^^x  +  y  =  14.-z, 
1.  Solve  the  system  <x-{-z  =  l  +  2y, 

[a;  4-2/  =15  -2  «. 

Solution  :  Rewriting  in  standard  form, 

3  a:  +  y,+  3  =  14, 
x-2y  -\-z  =  l, 
x+  ij  -\-2z  =  15. 


(1) 

(2) 
(3) 

(4) 
(«) 


LINEAR  SYSTEMS 


75 


From  I  and  II  preceding, 


14 
1 

1 
_  2 

1 

1 

15 

1 

2 

-26 

8 

1 

1 

-18 

1 

_  2 

1 

1 

1 

2 

2,  and  y  — 


3 

14 

1 

1 

1 

1 

1 

15 

2 

18 


-13 


8. 


The  value  of  s-can  now  be  more  easily  obtained  by  substituting 
the  values  of  x  and  y  already  found  in  (1),  (2),  or  (3)  than  by  means 
of  determinants. 

Substituting  in  (2),  2  +  ;^  =  1  +  6  ; 

whence  z  =  o. 

Check  :  (1)  +  (2)  +  (3)  gives 

5x  +  2?/  +  2;  =80  +  2?/-3  2:. 
'      Substituting,  10  +  6  +  5  =  30  +  6  -  15, 

or  21  =  21. 

(x  +  y  =  l^^-2z,      '  (1) 

2.  Solve  the  system  \  x  -\-  1  =  ^y,  •  (2) 

[x  +  4.z  =  -U.  (3) 

Solution :  Rewriting  in  standard  form  and  supplying  zero  coeffi- 
cients, 


X  +  ?/  -  2  ^  =  13, 
x  —  3y  —  0z=—7y 

X-Oy  +  4:Z=-14:. 


(4) 
(5) 
(6) 


Then 


18 

1 

_  2 

-7 

-3 

0 

-14 

0 

4 

1 

1 

-2 

1 

-3 

0 

1 

0 

4 

-44 
-22 


=  2. 


The  value  of  y  can  now  be  more  easily  obtained  by  substituting 
2  for  X  in  (2)  than  by  means  of  determinants. 

Accordingly  2+7  =  8?/; 

whence  ^      ,y  =  3. 

Similarly,  by  substituting  2  for  a:  in  (8),  2  +  4  2  =  —  14 ; 
whence  z  =—  4. 

Check  :   (1)  +  (2)  +  (8),  3  x  +  y -\- 4:z -^  7  =-1 -\- 2z  +  3  y. 
Substituting,  6  +  8-16  +  7=-l-8  +  9,  or0  =  0. 


► 


76  SECOND  COURSE  IN  ALGEBRA* 

As  we  have  seen,  determinants  have  a  very  useful  application  in 
the  solution  of  systems  of  linear  equations  in  two  or  three  varia- 
bles. With  some  practice  one  can  solve  such  equations  more  rapidly 
by  determinants  than  by  the  other  methods  which  have  been  given. 
If  the  student  studies  advanced  algebra,  he  will  learn  of  determi- 
nants of  the  fourth  and  higher  orders,  and  of  the  usefulness  of  sueh 
determinants  in  solving  linear  systems  in  four  or  more  variables. 
Moreover,  he  will  then  see  that  the  theory  of  determinants  is  an 
absolute  necessity  for  the  discussion  of  the  general  theory  of  linear 
systems  in  n  variables. 

EXERCISES 

Solve  for  x,  y,  and  z  as  in  the  two  preceding  examples : 

x-\-y-^z==l,  Ei^  =  Q 

^l,  x  +  y-z  =  2,  ^2        ' 

^  +  --1S 

ax  -{•  by  =  0, 

7.  ex  —  hz  =  2  be, 
bx  -\-  az  —  cy  =  b'^. 

lix  -{-  ky  —  lz  =  2  hJc, 

8.  Jcy  -  hx  -\-lz  =  2  Id, 
hx  —  ky  -\-lz  =  2  hi. 

mx  +  mx^  +  ^'2  =  Oj 

9.  7nx  -\-  x^-\-  mx,^  =  ma  —  a, 
tnx  —  3  mx^  -\-  x^'==  4:  'ma. 

Note.  Like  so  many  other  discoveries,  the  determinant  notation 
was  noticed  independently  by  two  men.  In  a  letter  to  a  friend,  writ- 
ten in  1693,  Leibnitz  outlined  the  method  of  solving  equations  by 
the  means  of  determinants ;  but,  so  far  as  we  know,  he  used  the  nota- 
tion in  his  own  work  very  little,  and  certainly  did  not  publish  it  dur- 
ing his  lifetime.  In  fact,  the  letter  in  which  this  reference  is  found, 
did  not  come  to  light  until  18.50,  and  the  fact  that  Leibnitz  knew 
anything  about  determinants  was  not  generally  recognized  until  after 
that  time. 


X  —  y  -\-  z  =  S. 

x  +  2y  +  z  =  l, 

2. 

2x-\-y-z  =  0, 

x-\,2y-\-z  =  0. 

2x-^y  =  5  +  z, 

3. 

x-2z  =  6, 

Sy  -{-  2z  =  x. 

^-\-y  =  % 

4. 

,x  +  z  =  2, 

2/ +  ^-  =  3. 

X  +  ?/  =  3  a, 

5. 

X  +  z  =  4a, 

?/.-f-  z  =  ^a. 

LINEAR  SYSTEMS  77 

In  1750  Cramer,  a  professor  in  the  university  at  Geneva,  rediscov- 
ered this  method  of  solving  linear  systems;  and  his  vrork  had  the 
good  fortune  to  be  accepted  by  scholars,  forming  the  real  beginning 
of  the  development  of  the  subject.  Since  that  time  a  great  many 
have  written  on  the  subject,  and  to-day  determinants  are  used  in 
every  field  of  advanced  mathematics. 


PROBLEMS 

^  1.  A  and  B  together  can  do  a  piece  of  work  in  3f  days.  If 
they  work  together  2  days  and  A  can  then  finish  the  job  alone 
in  2^  days  more,  how  many  days  does  each  require  alone  ? 

2.  A  man  and  a  boy  can  do  in  18  days  a  piece  of  work  which 
5  men  and  9  boys  can  do  in  3  days.  In  how  many  days  can 
one  man  do  the  work  ?  one  boy  ? 

3.  If  ax  +  hj  =  2  is  satisfied  by  a:;  =  2  and  y  =  S,  and  also 
by  X  =  6  and  y  =  5,  what  values  must  a  and  b  have  ? 

^  4.  A  launch,  whose  rate  in  still  water  is  12  miles  per  hour, 
goes  up  a  stream  whose  rate  is  2  miles  per  hour,  and  returns. 
The  entire  trip  requires  24  hours.  Find  the  number  of  hours 
required  for  the  trip  upstream  and  the  number  for  the  return. 

5.  Two  sums  are  put  at  interest  at  5%  and  6%  respec- 
tively. The  annual  income  from  both  together  is  $100.  If  the 
first  sum  had  yielded  1%  more  and  the  second  1%  less,  the 
annual  income  would  have  been  decreased  |2.    Find  each  sum. 

6.  A  sum  of  |4000  is  invested,  a  part  in  5  per  cent  bonds  at 
90,  and  the  remainder  in  6  per  cent  bonds  at  110.  If  the  total 
annual  income  is  $220,  find  the  sum  invested  at  each  rate. 

7.  A  train  leaves  M  two  hours  late  and  runs  from  M  to  P  at 
50%  more  than  its  usual  rate,  arriving  on  time.  If  it  had  run 
from  M  to  P  at  25  miles  per  hour,  it  would  have  been  48  minutes 
late.    Find  the  usual  rate  and  the  distance  from  M  to  P. 

8.  A  train  leaves  M  thirty  minutes,  late.  It  then  runs  to  ^ 
at  a  rate  20%  greater  than  usual,  and  arrives  6  minutes  late. 
Had  it  run  15  miles  of  the  distance  from  M  to  P  at  the  usual 


78  SECOND  COURSE  IN  ALGEBRA 

rate  and  the  rest  of  the  trip  at  the  increased  rate,  it  would 
have  been  12  minutes  late.  Find  the  distance  from  M  to  P  and 
the  usual  rate  of  the  train. 

9.  The  rate  of  a  passenger  train  is  62  feet  a  second  and  the 
rate  of  a  freight  train  38  feet  a  second.  When  they  run  on 
pamllel  tracks  in  opposite  directions  they  pass  each  other  in 
20  seconds.  The  length  of  the  freight  train  is  three  times  the 
length  of  the  passenger  train.    Find  the  length  of  each. 

10.  The  rate  of  a  passenger  train  is  45  miles  per  hour  and 
the  rate  of  a  freight  train  is  30  miles  per  hour.  The  freight 
train  is  240  feet  longer  than  the  passenger  train.  When  the 
trains  run  on  parallel  tracks  in  the  same  direction  they  pass 
each  other  in  1  minute  and  20  seconds.  Find  the  length  of 
each  train. 

^  11.  The  length  of  a  freight  train  is  1430  feet  and  the  length 
of  a  passenger  train  550  feet.  When  they  run  on  parallel  tracks 
in  opposite  directions  they  pass  each  other  in  18  seconds,  and 
when  they  run  in  the  same  direction  they  pass  each  other  in  1 
minute  and  30  seconds.    Find  the  rates  of  the  trains. 

12  Two  contestants  run  over  a  440-yard  course.  The  first 
wins  by  4  seconds  when  given  a  start  of  2(^0  feet.  They  finish 
together  when  the  first  is  given  a  handicap  of  40  yards.  Find 
the  rate  of  each  in  feet  per  second. 

13.  It  is  desired  to  have  a  10-gallon  mixture  of  45%  alcohol. 
Two  mixtures,  one  of  95%  alcohol  and  another  of  15%  alcohol, 
are  to  be  used.  How  many  gallons  of  each  will  be  required  to 
make  the  desired  mixture  ? 

14.  A  chemist  has  the  same  acid  in  two  strengths.  Eight 
liters  of  one  mixed  with  12  liters  of  the  other  gives  a  mixture 
84%  pure,  and  3  liters  of  the  first  mixed  with  2  liters. of  the 
second  gives  a  mixture  86%  pure.  Find  the  per  cent  of  purity 
of  each  acid. 

15.  The  crown  of  Hiero  of  Syracuse,  which  was  part  gold 
and  part  silver,  weighed  20  pounds,  and  lost  li  pounds  when 


LINEAR  SYSTEMS  79 

weighed  in  water.  How  much  gold  and  how  much  silver  did  it 
contain  if  19^  pounds  of  gold  and  10^  pounds  of  silver  each 
lose  one  pound  when  weighed  in  water  ? 

^  16.  One  angle  of  a  triangle  is  twice  another,  and  their  sum 
equals  the  third.  Find  the  number  of  degrees  in  each  angle  of 
the  triangle. 

^17.  The  sum  of  three  numbers  is  108.  The  sum  of  one  third 
the  first,  one  fourth  the  second,  and  one  sixth  the  third  is  25. 
Three  times  the  first  added  to  four  times  the  second  and  six 
times  the  third  is  504.    Find  the  numbers. 

18.  The  sum  of  three  numbers  is  217.  The  quotient  of  the 
first  by  the  second  is  5,  which  is  also  the  quotient  of  the  sec- 
ond by  the  third.    Find  the  numbers. 

d  19.  If  the  tens'  and  units'  digits  of  a  3-digit  number  be 
interchanged,  the  resulting  number  is  27  less  than  the  given 
number.  If  the  same  interchange  is  made  with  the  tens'  and 
hundreds'  digits,  the  resulting  number  is  180  less  than  the 
given  number.   The  sum  of  the  digits  is  14.   Find  the  number. 

20.  In  one  hour  a  tank  which  has  three  intake  pipes  is  filled 
seven  eighths  full  by  all  three  together.  The  tank  is  filled  in 
1^  hours  if  the  first  and  second  pipes  are  open,  and  in  2  hours 
and  40  minutes  if  the  second  and  third  pipes  are  open.  Find 
the  time  in  hours  required  by  each  pipe  to  fill  the  tank. 

21.  The  sum  of  two  adjacent  sides  of  a  quadrilateral  is  140 
inches.  The  sum  of  the  first  of  these  and  the  side  opposite  is 
160  inches ;  the  sum  of  the  first  side  and  the  fourth  side  is  172 
inches.    The  perimeter  is  352  inches.    Find  each  side. 

22.  The  sum  of  two  sides  which  meet  at  one  of  the  vertices 
of  a  quadrilateral  is  20  feet.  The  sum  of  two  which  meet  at 
the  next  vertex  is  27  feet.  The  sums  of  the  two  pairs  of  oppo- 
site sides  are  23  feet  and  29  feet  respectively.   Find  each  side. 

23.  The  sums  of  three  pairs  of  adjacent  sides  of  a  quadri- 
lateral are  respectively  80  feet,  108  feet,  and  116  feet.    The 


«0  SECOND  COURSE  IN  ALGEBRA  J 

difference  of  the  fourth  pair  of  adjacent  sides  is  24  feet.    Fina 
each  side. 

24.  Two  chairs  cost  li  dollars.  The  first  cost  m  cents  more 
than  the  second.    Find  the  cost  of  each  in  cents. 

25.  A  and  B  together  have  d  dollars.  A  gives  c  dollars  to 
B,  after  which  B  gives  m  dollars  to  A.  Then  A  has  \  as  many 
dollars  as  B.    Find  the  number  of  dollars  each  had  at  first. 

26.  A  and  B  together  can  do  a  piece  of  work  in  m  days.  B 
works  c  times  as  fast  as  A.  How  many  days  does  each  require 
alone  ? 

27.  A  man  rows  m  miles  downstream  in  t  hours  and  returns 
in  a  hours.    Find  his  rate  in  still  water  and  the  rate  of  the  river. 

28.  A  man  dying  leaves  a  widow  and  eleven  children.  The 
law  provides  that  the  widow  shall  receive  one  half  of  the  estate 
and  that  the  other  half  shall  be  divided  equally  among  the 
children.  The  executor  of  the  estate,  after  paying  all  debts, 
has  |3400  in  cash.  But  five  of  the  children  had  borrowed  from 
their  father  |400  each,  for  which  he  had  accepted  their  notes. 
The  executor  found  these  notes  worthless.  How  shall  he  divide 
the  cash  on  hand  ? 

0  29.  Solve  in  positive  integers  5  ic  +  2  ?/  =  42. 

Hint,  x  = — ^  =  8  +  — — ^ . 

5  5 

2  —  2  « 

Now  if  X  is  to  be  integral, must  be  integral ;  that  is,  2  —  2  ?/ 

5 

must  be  an  integral  multiple  of  5.    Hence  the  least  value  of  y  is  G. 

The  various  related  sets  of  values  which  satisfy  this  equation 
may  be  effectively  represented  to  the  eye  by  the  graph  of  the 
equation.  Then  if  the  line  whose  equation  is  5x  -\-  2j/  =  4:2 
passes  through  any  points  both  of  whose  coordinates  are  posi- 
tive integers,  each  pair  of  these  values  is  a  set  of  roots.  If  the 
line  does  not  enter  the  first  quadrant,  we  can  see  at  a  glance 
that  the  equation  has  no  set  of  roots  which  are  positive 
integers. 


LmEAR  SYSTEMS  81 

30.  Solve  in  positive  integers  7ic  +  2?/  =  36,  and  illustrate 
the  result  graphically. 
^31.  In  how  many  ways  can  a  debt  of  |73  be  paid  with  five- 
dollar  and  two-dollar  bills  ?     Illustrate  the  result  graphically. 

32.  A  man  buys  calves  at  |6  each  and  pigs  at  |4  each, 
spending  |72.    How  many  of  each  did  he  buy  ? 

33.  In  how  many  ways  can  $1.75  be  paid  in  quarters  and 
nickels  ?  * 

34.  A  farmer  sells  some  calves  at  |6  each,  pigs  at  |3  each, 
and  lambs  at  |4  each,  receiving  for  all  $126.  In  how  many 
ways  could  he  have  sold  32  animals  at  these  prices  for  the 
same  sum  ?    Determine  the  various  groups. 

y,    35.  In  how  many  ways  can  a  sum  of  $2.40  be  made 'up  with 
nickels,  dimes,  and  quarters,  on  the  condition  that  the  number 
of  nickels  used  shall  equal  the  number  of  quarters  and  dimes 
together  ?    Determine  the  various  groups. 
• 


CHAPTER  VI 

ROOTS,  RADICALS,  AND  EXPONENTS 

48.  Definitions.  The  square  root  of  any  number  is  one  of  the 
two  equal  factors  whose  product  is  the  number. 

For  a  given  index  the  principal  root  of  a  number  is  its  one 
real  root,  if  it  has  but  one ;  or  its  positive  real  root  if  it  has 
two  real  roots. 

From  the  law  of  signs  in  multiplication  it  follows  that 

Every  positive  number  or  algebraic  expression  has  two  square 
roots  which  have  the  same  absolute  value  but  opposite  signs. 

It  is  customary  to  speak  of  the  positive  square  root  of  a 
number  as  the  principal  square  root;  and  if  no  sign  precedes 
the  radical,  the  principal  root  is  understood.  When  both  the 
positive  and  the  negative  square  roots  are  considered,  both 
signs  must  precede  the  radical. 

For  extracting  the  square  roots  of  any  monomial  we  have  the 

E-ULE.  Write  the  square  root  of  the  numerical  coefficient 
preceded  by  the  sign  ±  and  followed  by  all  the  letters  of  the 
tnonomial,  giving  to  each  letter  an  exponent  equal  to  one  half 
its  exponent  in  the  monomial. 

A  rule  much  like  the  preceding  one  holds  for  fourth  root, 
sixth  root,  and  other  evei^  roots. 

The  even  roots  of  negative  numbers  are  considered  in  the  chapter 
on  Imaginaries. 

In  this  chapter  only  a  single  odd  root  of  a  number  is  con- 
sidered; that  is,  the  principal  odd  root. 

The  cube  root  of  any  number  is  one  of  the  three  equal  factors 
whose  product  is  the  number. 

82 


ROOTS,  RADICALS,  AND  EXPONENTS  83 

For  extracting  the  cube  root  of  a  monomial  we  have  the 
Rule.    Write  the  cube  root  of  the  numerical  coefficient  followed 
hy  all  the  letters  of  the  Ttionomial,  giving  to  each  letter  an  expo- 
nent fiqual  to  one  third  of  its  exponent  in  the  monomial. 

A  rule  much  like  the  preceding  one  holds  for  fifth  root, 
seventh  root,  and  other  odd  roots. 

49.  Square  root  of  polynomials.  Extracting  the  square  root 
of  a  number  is  essentially  an  undoing  of  the  work  of  multipli- 
cation.   The  square  of  any  polynomial  may  be  represented  by 

(7t  +  f  +  tif  =  Ji"  +  2  A^^  +  ?^  +  2  hu  -\-2tu  +  u^ 
=  h^^{2h  +  t)t+{2h  +  2t  +  u)u. 

A  little  study  of  this  last  form,  and  a  comparison  with  the 
example  which  follows,  will  make  clear  the  reason  for  each 
step  of  the  process. 

EXAMPLE 

7*2  +2M+t^+2  hu  +  2  m  +  u-\h  +  /  +  M 
7/2 
First  trial  divisor,  2  h 


First  complete  divisor, 

Second  trial  divisor,  21i  -\-  2t 
Second  complete  divisor, 


+  2Tit  +  fi      . 
■\-2ht-\-  t^  =  (2h  +  t)t 


2h-{-2t+  u 


+  2hu  +  2  til  +  u^ 

+  2  /m  +  2  tu  +  m2  =  (2h  +  2t  +  u)  u 


Therefore  the  required  roots  are  ±  (A  +  ^  +  «)• 
EXERCISES 

1.  State  the   rule  for  the   sign  of   (a)   the   odd  root  of  a 
number;  (b)  the  even  root  of  a  number. 

2.  State  the  rule  for  extracting  the  fourth  root  of  a  monomial. 

3.  State  the  rule  for  extracting  the  fifth  root  of  a  monomial. 

4.  How  can  one  obtain  the  fourth  root  of  a  polynomial  ? 

5.  State    the    rule    for    extracting    the    square    root    of    a 
polynomial. 


84 


SECOND  COURSE  IN  ALGEBRA 


6.  Can  one  obtain  the  fifth  root  of  a  number  (a)  by  ex- 
tracting the  square  root  of  its  cube  root  ?  (b)  by  extracting 
the  cube  root  of  its  square  root?    Explain. 

Extract  the  square  roots  of : 

7.  X*  +  ^x^  -  2 x^  -  12 X  +  9. 

8.  a«-10a*-4a«  +  25a2_f-20«^4-4. 
.  9.  4a«-fl2a*-7-24a-*H-16a-«. 

10.  49 c-«  _  28 C-*  +  74 c-2  _  20  -f-  25 c\ 

11.  9x^-exi  -i-x^-eGx^  -\-22x^  +  121x. 

12.  25ic«-10a'2  +  90;r^  +  a^-18a^^  +  81xi 

.13.  16  W-'  _  8  m-*  +  104m  -  26m*  +  169 m^  +  m-\ 
14.  I  a*  -  2a«  +  7|  a^  -  -s^^-a  +  V". 


15. 
16. 


9  c' 

1 

4:a^ 


'    ,   16^  ,   4a 


3c 


16c      20 

a         3 


ct' 


6a        1 
5        5x 


Find  the  first  four  terms  in  the  square  root  of : 

17.  l  +  2x.  18.  -2/  +  ^'- 

19.  Find  the  first  three  terms  in  the  fourth  root  of  the 
expression  in  (a)  Exercise  7 ;  (h)  Exercise  11. 

50.  Graphical  method  of  extracting  roots.  When  obtaining  the 
square  roots  of  arithmetical  numbers  by  the  graphical  method 
we  proceed  as  follows  :  Let  x  represent  any  number  and  y  the 
square  of  that  number ;  that  is,  we  let  y  =  x\  Then  we  con- 
struct the  graph  of  this  equation,  obtaining  first  the  table : 


X 

i 

1 

2 

3 

4 

5 

6 

7 
49 

8 

y 

i 

1 

4 

9 

16 

25 

36 

64 

Plotting  these  values,  we  obtain  curve  OD  of  page  85. 
From  this  curve  we  can  read  off  the  square  root  of  any 
number  between  1  and  100  correct  to  one  decimal  j)lace. 


ROOTS,  RADICALS,  AND  EXPONENTS 


85 


^ 

00 

A 

C 

[ 

j 

— 

ac 

j 

j 

90 

1 

1 

j 

j 

1 

-80 

-7  5 

1 

-70 

1 

/ 

1.5 

/ 

60 

;■ 

1 

i 

.... 

J 

I 

k\ 

f=^ 

1 

: 

cl 

"> 

Ai 

/ 

X 

7 

'■ 

/ 

V 

/ 

1 

/ 

-a  j; 

j 

r 

j 

/ 

-3  0 

/ 

/I 

0  A. 

/ 

J 

/ 

/ 

y 

J- 

i 

/ 

- 

■IQ 

1 

/ 

j 

/ 

/ 

/ 

/ 

y 

{^ 

y 

-O 

^ 

^ 

ifj 

^ 

Bf 

...s.. 

i 

1 

0 

-X 

bG  SECOND  COURSE  IN  ALGEBRA 

Cui-ve  OA  is  a  portion  of  the  graph  ofy  =  x^,  and  5C  is  a 
continuation  of  OA.  From  this  curve  we  can  obtain  the  cube 
root  of  any  number  between  1  and  200  correct  to  one  decimal 
place.  The  cube  root  of  numbers  between  1  and  100  we  obtain 
from  curve  OA ;  for  numbers  between  100  and  200  we  obtain 
the  cube  root  from  BC. 

If  one  desires  greater  precision  or  a  larger  range  of  numbers,  or 
both,  he  can  obtain  them  by  using  a  large  piece  of  cross-section  paper 
and  a  different  scale.  Such  a  curve,  if  carefully  drawn,  is  convenient 
for  any  computation  not  requiring  great  accuracy.  The  point  in  its 
favor  is  that  one  can  read  off  the  square  roots  or  the  cube  roots  more 
rapidly  than  he  can  obtain  them  by  the  methods  of  §§  50  and  142,  or 
even  by  logarithms. 

The  graphical  method  can  also  be  used  to  extract  fourth  and 
higher  roots. 

EXERCISES 
From  the  graph  read  : 

1.  The  square  root  of  (a)  20 ;  (h)  45 ;  (c)  59 ;  (d)  68. 

2.  The  cube  root  of  (a)  25  ;  (b)  18  ;  (c)  52 ;  (d)  165  ;  (e)  150. 

3.  The  square  of  (a)  2.4 ;  (b)  6.1 ;  (c)  7.9 ;  (d)  8.3. 

4.  The  cube  of  (a)  3.2;  (b)  3.9;  (c)  2.8;  (d)  5.6;  (e)  4.8. 

51.  Square  root  of  arithmetical  numbers.  The  abbreviated  proc- 
ess of  extracting  a  square  root  of  an  arithmetical  number  is. 
as  follows : 

7'32'67'89  [2706^8  + 
4 


47 


332 
329 


5406 
54128 


36789 
32436 


435300 
433024 


2276 
Therefore  the  square  roots  of  7326789  are  ±  2706.8 


ROOTS,  RADICALS,  AND  EXPONENTS  87 

It  follows  from  the  preceding  example  that  the  work  of  extracting 
the  positive  square  root  of  a  number  may  be  a  never-ending  process. 
The  number  7,326,789  has  no  exact  square  root ;  and  no  matter  how 
far  the  work  is  carried,  there  is  no  final  digit.  As  the  work  stands 
we  know  that  the  required  root  lies  between  2706.8  and  2706.9.  It 
is  correct  to  say  that  2706.8  is  approximately  the  square  root  of 
7,326,789,  or  that  it  is  the  square  root  correct  to  five  figures. 

The  method  just  illustrated  for  extracting  the  positive  square 
root  of  a  number  is  the  one  commonly  used.    For  it  we  have  the 

Rule.  Begin  at  the  decimal  point  and  point  off  as  many 
pjeriods  of  two  digits  each  as  possible  :  to  the  left  if  the  numhei' 
is  an  integer,  to  the  right  if  it  is  a  decimal ;  to  both  left  and 
right  if  the  number  is  part  integral  and  part  decimal. 

Find  the  greatest  integer  whose  square  is  equal  to  or  less  than 
the  left-hand  period,  and  ivrite  this  integer  for  the  first  digit  of 
the  root. 

Square  the  first  digit  of  the  root,  subtract  its  square  from  the 
first  period,  and  annex  the  second  period  to  the  remainder. 

Double  the  part  of  the  root  already  found  for  a  trial  divisor, 
divide^  it  into  the  remainder  (omitting  froiYi  the  latter  the  right- 
hand  .digiC),  and  write  the  integral  part  of  the  quotient  as  the 
next  digit  of  the  root. 

Annex  the  root  digit  just  found  to  the  trial  divisor  to  mahe 
the  complete  ;divisor,  multiply  the  complete  divisor  by  this  root 
digit,  subtract^  the  result  from  the  dividend,  and  annex  to  the 
remainder  the  'ii^xt  period,  thus  making  a  new  dividend. 

Doidjle  the  part  of  the  root  already  found  for  a  new  trial 
divisor  and  proceed  as  before  until  the  desired  number  of  digits 
of  the  root  have  been  found. 

After  extracting  the  square  root  of  a  number  involving  deci- 
mals, point  off  one  decimal  place  in  the  root  for  every  decimal 
period  in  the  number. 

Check.  If  the  root  is  exact,  square  it.  The  result  should  be  the 
original  number.  If  the  root  is  inexact,  square  it  and  add  to  this 
result  the  remainder.    The  sum  should  be  the  original  number. 


88  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 
Find  the  square  roots  of : 

1.  6889.  3.  .6724.  5.  4.2025. 

2.  56169.  4.  1.4641.  6.  .04028049. 
Extract  the  square  roots,  correct  to  four  decimal  places,  of : 

7.  5.  8.  .07.  9.  V.  10.  ^^-. 

11.  Find  the  hypotenuse  of  the  right  triangle  whose  legs 
are  183  and  264  respectively. 

12.  A  baseball  diamond  is  a  'square  90  feet  on  each  side. 
Find  the  distance  from  the  home  plate  to  second  base,  correct 
to  .01  of  a  foot. 

13.  The  hypotenuse  of  a  right  triangle  is  207  feet  and  one 
leg  is  83  feet.    Find  the  other  leg,  correct  to  .01  of  a  foot.    , 

14.  The  hypotenuse  and  one  leg  of  a  right  triangle  are 
respectively  292849  and  207000.   Find  the  other  leg. 

15.  The  side  of  an  equilateral  triangle  is  11  inches.  Find  its 
altitude,  correct  to  .1  of  an  inch. 

16.  Find  the  side  of  an  equilateral  triangle  whose  altitude 
is  10  inches,  correct  to  .001  of  an  inch. 

17.  Find  the  area  of  a  triangle  whose  sides  are  12,  27,  and 
35  inches  respectively,  correct  to  .001  of  a  square  foot. 

Fact  from  Geometry.  If  a,  h,  and  c  represent  the  sides  of  a  triangle, 
and  s  equals  one  half  of  a  +  h  ■\-  c,  the  area  of  the  triangle  equals 
■yjs  (s  —  a)  (s  —  h)  {s  —  c). 

18.  By  the  method  of  Exercise  17  find,  correct  to  .01  of  a 
square  inch,  the  area  of  a  triangle  each  side  of  which  is  22  inches. 

19.  Find  correct  to  two  decimals  the  sum  of  all  of  the  diag- 
onal lines  that  can  be  drawn  on  the  faces  of  a  cube  whose  edge 
is  9  inches. 

20.  Find  the  radius  of  a  circle  whose  area  is  40  square  feet. 

21.  Find  the  diagonal  of  a  room  whose  dimensions  in  feet 
are  14,  20,  and  30. 


ROOTS,  RADICALS,  AND  EXPONENTS  89 

22.  Find  the  diagonal  of  a  cube  whose  edge  is  1  foot. 

23.  A  room  is  24  feet  by  40  feet  by  14  feet.  What  is  the 
length  of  the  shortest  broken  line  from  one  lower  corner  to 
the  diagonally  opposite  upper  corner,  the  line  to  be  in  part  on 
the  walls  or  the  floor,  but  not  through  the  air  ? 

24.  Take  any  two  integers  and  form  three  others  from  them 
thus :  find  the  sum  of  their  squares,  the  difference  of  their 
squares,  and  twice  their  product.  Is  the  square  of  one  of  the 
three  resulting  numbers  equal  to  the  sum  of  the  squares  of  the 
other  two  ?  Discuss  this  with  reference  to  the  sides  of  a  right 
triangle. 

25.  One  leg  of  a  right  triangle  is  28.  Find  integ^-al  values 
for  the  other  two  sides. 

52.  Fundamental  laws  of  exponents.    The  laws  of  exponents 
may  be  stated  as  follows  : 
I.  Law  of  Multiplication, 

Law  I  may  be  stated  more  completely  thus : 

y.a  ,  y 6  ,  yC  ^^  ^  __  y.a  +  &  +  c+ •  •  • 

This  follows  directly  from  the  definition  of  an  exponent  and 
from  the  Associative  Law.  For  instance,  xx  =  x^,  and  xxx  =  x^,  and 
xxxxx  =  x^  by  definition.    Hence  xx  ■  xxx  =  xxxxx,  or  x^  •  x^  =  x^. 

IL  Law  of  Division, 

jir° -5-Jl:^  =  JC°-^• 
This  follows  from  Law  I.    For  by  that  law  x"  =  a:«  -  ^  •  a:*.    Hence, 
dividing  both  sides  of  this  equation  by  x*,  we  have  x"-  -^  a*  =  ^«  -  ^. 

IIL  Law  of  Involution,  or  raising  to  a  power, 

(x°)^  =  x''K 

This  follows  from  Law  I,  when-instead  of  the  distinct  factors  a;«, 
x^,  and  X'',  etc.,  we  have  b  factors,  each  equal  to  x«. 

}•.     Law  III  includes  the  more  general  forms 

I  (x°y^y  =  a:V^  and  (((x°yy)  •  •  •  =  Jtr'^^^-. 


90  SECOND  COURSE  IN  ALGEBRA 

IV.  Law  of  Evolution,  or  extraction  of  roots, 

hr—  - 


It  is  assmned  that  these  laws  hold  for  all  real  values  of  a,'i&, 

and  c,  excepting  under  IV,  where  h  cannot  be  zero.  \ 

\ 
53.  Zero  and  negative  exponents.    The  meaning  of  a  zero  ex3 

ponent  and  of  a  negative  exponent  was  explained  on  page  6. 

It  follows  from  the  meaning  of  a  negative  exponent  there 
explained  that 

Any  factor  of  the  numerator  of  a  fraction  rnay  be  taken 
from  the  numerator  and  written  as  a  factor  of  the  denominator^ 
and  vice  versa,  if  the  sign  of  the  exjjonent  of  the  factor  he 


54.  Meaning  of  a  fractional  exponent.  The  fourth  law  ex- 
presses the  meaning  of  a  fractional  exponent.  The  meaning 
may  be  made  clearer  thus  : 

From  Law  I,  x^  ■  x-  =  x.    Since  x^  multiplied  by  itself  gives 

£c,  x^  is  another  way  of  writing  "V^ ;  that  is,  x^  =  V^. 

Similarly,  x^  -x^  -x^  =^  x.     Therefore  a:;*  is  another  way  of 

writing  V^ ;  that  is,  x^  =  s/x. 
1 
In  general  terms  x"  is  another  way  of  writing  -y/x  -,  that  is, 

x^  =  -^. 

Further,  x^  =  x^  -x^  -x^  =  (x^f  =  (  Vxf, 

and  x^=x^'^  =  (x^)  ^  =  VS. 

Hence  (  Vxf  =  Vx^. 

Similarly,  x^  =;)c^  -x^  =  {x^f  =  (  V^)^ 

and  V?  =  (x^y  =  (x  ■  x)^  =  x/^  -x^  =  x^. 

Therefore  ( -y/xf  =  V^. 

a 

In  more  general  terms  x"  =  ( "v^)''  =  '\/x°. 


ROOTS,  RADICALS,  AND  EXPONENTS  91 

EXERCISES 

Write  with  positive  exponents  and  then  simplify  results : 

^1.  83.  3.  32*  *^5.  2-\  7.  (f)-"^. 

2.   16l  rCl25l  6.  49"  i  s:  (f)-l 

<^9.  8-3.  12.  (32)-^  ^^    4 -2^ -2'^-^ 

10.  (-27)*.        t/13.  16-H  '      8-2^-" 

Jll.  (.09)1  14.  (.008)-^  16.  (§)-'•  (I)' -21 

^77,  5  .  2«  -  1^  +  (5  •  2)0  -  2(1)'  -  (8-3)  +  (9)-^ 

2-^  —  3-^  2-1-8-1      ^-¥ 

18.  — ^-^ Hint.         ^_,        =  -^-  =  etc. 

4-2  _  3-1  23.  :z;-\  *      „„    T?/^-^.^-* 

19.  .    .,   ■   ^    .'  ^    28.      _.,_,.      , 


4-'^ +  3-1 
3-1  +  4- 
3-^  +  4- 


24.  mc-^ 
20.  Tl^t'l '        "^  25.  3 ..-  V.  1/29.  ?^ 


♦'^     4"  3  _  2 


4x 


21.    .    .    .    ^   .'  *""•    m-3  30. 


22    3^-3-3^  +  ^  27    6x^^  ^^    4-%%?/ 


«u. 

m-' 

27. 

6  x"  ^??? 

9 

a-^  +  x-^ 

35. 

X-' 

x-"-  +  a-'' 

'id 

x-''  -  a-'' 

4-1  +  2-1     ,  '"^  amc- 

•  3-3 

9-3^  +  ' 

2 
32.  — -r^. z  •      Hint.   — -^ =  .    "   .    =  etc. 


:c-i  — a-1  a;-^  +  a-^  .-      a-^  +  Sa:;-'^ 

34.-^—..  36.-^-..         38.^"*  +  '^''  +  ! 


a^-2  _aj-i  _^i 


Write  without  a  denominator  : 


39.  2!.  U   ,1.  1|L.  43.       1«--^ 


a  a~'^c^  m  (a  —  Trif 


j,^    4a"^  ,^       me 

40.  — r-  42. 44. 


2^3 


cm^  m  +  c  mr'^ar  (m  —  a)- 


92 


SECOND  COURSE  IN  ALGEBRA 


Write,  using  positive  fractional  exponents  instead  of  radical 
signs : 

45.  V^.  49.  -^\  53.   V2-i-a. 

46.  -^"c.  50.  4  ^a\  (^4.   VsT^. 

47.  ^t  -'''\  CU^3  -v^^  55.   -v^4H-a---^. 

48.  ^.    '  '^^  52.  5cV^.  56.  Sc^^h-^^-^ 

57.    V  a  V^.    Solution  :   V a  Va  =  v  a  •  a^  =  V  a^  =  a*. 


58. 


Vm. 


59. 


Vc 


60.  4v^^8. 


61.   ^c-¥c. 


62.  6V75V9. 


MISCELLANEOUS  EXERCISES 
Perform  the  indicated  operations  : 
-v  1.  x^-x-\  13.  {{xy)^^. 

2.  x^  ,x^.  14.  (??i2)2«. ' 

3.  x^-x^.  ^  15.  (ic-i)-2'^.    ■ 

16.  (xy-^^. 

17.  (a;"-3)»  +  l 


4.  e'-e-^. 

5.  e^T^-e^-^" 

6.  e«-^-e^-«. 

7.  (e^)^ 
V   8.  (e^)«. 

9.  W. 


18.  ((a;-3y«)-^«. 

1 

19.  (£c«'-«')«-^ 

20.  [(4ic«/.8.42]i      32.  8^«  h- 


24.  x^  -T-  x^. 

25.  e^H-e-^ 

26.  e«-i^el 

27.  e«'-i--e«- 

28.  42-2'^. 
V29.  82^4^ 

30.  2^42^8'-^. 

31.  4^"  H-  2^ 


^^^. 


10.  {e^)\ 


ii.  (.0- 

12.  (el) 


21.  a;^  •  a;^  -f-  a?^  33.  4^''  •  2^"  -=-  82^ 

'^  22. '  ic«  •  x»  --  a;!^  34.  {x^  -2x- y. 

23.  x^  -^  xK  35:  (x*  -  3  x^y. 

'^  36.  (a2«-2  -  2a«-i  -  8)(a«-i  +  4.)(a^-^  -  2). 

37.  (4  e^  +  e-^  -  4)^.  41.  (a;  -  y^) -^ (x^  -  i/). 

38.  (2a-3«-i-4a-y.    42.  {a-^-x-^)^{x-j -\- a-^). 

39.  (ic^  _  ^2>)  ^  (^^  _f_  ^^i)_  N43,  (^-  8  ^  ^-  3N)  ^  (;^-  f  _|.  ^-  f ). 

40.  (a-^-2)-i-(a;-2*).        44.  (Sa--^- 48  a*) --(x"^  +  2  r.). 


ROOTS,  RADICALS,  AND  EXPONENTS  93 

45.  (x"  -  if)  ^  {xi  +  xij^  +  x^'y  +  y^)  H-  {x^  +  y^). 

46.  {x  +  32  a)  -^  (x*  +  2  a' )  +  2  x^a^  +  8  ^-^a*. 

47.  ((^2n  +  2  _  ^^«  +  i  _  i2)(4  -  a'^  +  i)  +  48  +  16  a"+i. 

/^6x  _  -j  Q     2x  _   Q\ 

-48.  (^  -  ^^^)  ^  {e^  -e-^)  +  {e^  +  .-^l 

49.  (e5«^-5e^^  +  10e^-10e-^4-5e-3^-e-^") 

L_(e2^ +  6-2^-2). 

50.  Point  out  the  error  in  the  following : 
Let  X  be  a  number  such  that  e^  =  —  1. 
Then  e'^  =  l. 

Hence  2  ic  =  0,  or  £c  =  0, 

and  e^  =  e^  =  1. 

Therefore  1  =  -  1. 

55.  Classification  of  numbers.  All  the  numbers  of  algebra  are 
in  one  or  the  other  of  two  classes,  real  numbers  and  imaginary 
numbers. 

Eeal  numbers  are  of  two  kinds,  rational  numbers  and  irra- 
tional numbers.  C 

A  rational  number  i^  a  positive  or  a  negative  integer,  or  a 
number  which  may  be  expressed  as  the  quotient  of  two  such 
integers. 

Any  real  number  which  is  not  a  rational  number  is  an  irra- 
tional number. 

A  pure  imaginary  number  is  the  indicated  even  root  of  a 
negative  number. 

56.  Radicals.    A  radical  is   an  indicated  root   of  the   form 
.  —  11 

■\/?i  or  c  -Vn,  or  of  the  form  n""  or  en''. 

A  surd  is  an  irrational  root  of  a  rational  number. 

The  index  determines  the  root  to  be  extracted  and  the  order 
of  the  radical. 

The  radicand  is  the  nmnber,  or  expression,  under  the  radical 
sign. 


94         SECOND  COURSE  IN  ALGEBRA 

EXERCISES 
Write  with  radical  signs : 

1.  xl  4.  4£ci  7.  Sj(c^x)i 

2.  (ac)i  5.  6cx^.  4  a^  (c  —  x)^ 

3.  (4ic)i  6.  Al 

Find  the  numerical  value  of : 
9.  4i  13.  4i 

10.  36^.  14.  27l 

11.  64i  15.  (-  8)1 

12.  81^.  16.  (^)i 

Write  with  fractional  exponents  : 

21.  ^\  25.  5-\^S?. 

22.  ^4?.  26.  6^64^.  ■"'        -v^5^V^ 

23.  SV^^  27.  5  ^- 125  c^.  30.  -^^--s/c. 

24.  4  \/4^.  28.  c  V(.x  4-  «)'•  31.   V^ 


o. 

J(c-x) 

17. 
18. 
19. 
20. 

(-  32)1 

(-125)t.GV)t. 
(-  243)^  •  (81)1 

oo 

5  c  Vaic^  •  -y/d  x^ 

„2a 


32.  Give  an  example  of  (a)  a  real  number ;  (b)  an  imaginary 
nmnber ;  (c)  a  rational  number ;  (c?)  an  irrational  nmnber ;  (e)  a 
radical ;  (/)  a  surd ;  (g)  an  index ;  (h)  a  radicand ;  (i)  tlie  principal 
odd  root  of  a  positive  number ;  ( /)  the  principal  even  root  of  a 
positive  number ;  (A:)  the  principal  odd  root  of  a  negative  number. 

33.  What  is  the  distinction  between  a  rational  number  and 
an  irrational  one  ? 

34.  Which  of  the  numbers  8,  f ,  .343,  Vi,  V3,  and  tt 
(tt  =  3.14159  +)  are  rational  ?    irrational  ? 

35.  Give  a  geometrical  illustration  of  an  irrational  number 
by  means  of  a  right  triangle. 

36.  Is  a  radical  always  a  surd  ?    Illustrate. 

37.  Is  a  surd  always  a  radical  ?    Illustrate. 

38.  Distinguish  between  a  surd  and  a  radical. 


ROOTS,  RADICALS,  AND  EXPONENTS  95 


39.  Which  of  the  numbers  V3,  V4,  -^27,  Vv'e,  V2  +  Vs, 
and  Vtt  are  surds  ?   Which  are  radicals  ? 

40.  What  is  the  principal  root  of :   V 4,  Vs,  and  V  —  8  ? 

41.  Name  the  order  of:   Vo,  d^,  "v5,  c%  and  VW. 

42.  How  many  real  numbers  can  be  found  for  a  designated 
odd  root  of  (a)  a  positive  real  number  ?  (b)  a  negative  real 
number  ? 

43.  Change  the  word  "  odd  "  in  («)  of  Exercise  42  to  "  even/^ 
and  answer. 

57.  Simplification  of  radicals.  The  form  of  a  radical  expres- 
sion may  be  changed  without  altering  its  numerical  value.  It 
is  often  desirable  to  change  the  form  of  a  radical  so  that 
its  numerical  value  can  be  computed  with  the  least  possible 
labor. 

The  simplification  of  a  radical  is  based  on  the  general 
identity : 

A  radical  is  in  its  simplest  form  when  the  radicand 
(«)  7s  integral. 

(b)  Contains  no  rational  factor  raised  to  a  power  which  is 
equal  to,  or  greater  than,  the  order  of  the  radical. 

(c)  Is  not  raised  to  a  power,  unless  the  exponent  of  the  power 
and  the  index  of  the  root  are  prime  to  each  other. 

Eor  the  meaning  of  (a),  (h),  and  (c)  study  carefully  the 

EXAMPLES 

Of  {a)  :  1.  Vl  =  \/i  =  VT^  =  Vl  V6  =:  i  V6. 

3  \lBx 


96 


SECOND  COURSE  IN  ALGEBRA 


2.  5^^/2^'=  5  y/Sx^ ' 3x'=  5-s/(2xy  ■  Sx'  =  10x-V3^^. 

3.  Vl6  -  8  V2  =  V4 (4  -  2  V2)  =  2  V4  -  2  V2. 
Of  (c)  :   1.  ^  =  -v^  =  2^  =  2i  =  V2. 

2.  ^  =  ^  =  3t  =  3^  =  ^. 

3.  \^7HJ^  =  Jb^  =  ah  =  bVa. 


EXERCISES 

Express  in  simplest  form : 

1.  Vi8.         10.  vi- 


2.  7l6. 

3.  2V75. 

4.  4  ^/-  54. 

5.  4V4O. 

6.  \Qt. 
(7.  \^3V9. 

8.  v^. 

Express  entirely  under  the  radical  sign : 


11.  v^- 

12.  rh_ 

13.  6^-i. 

14.  vi-a)^ 

15.  V3'-©'. 

le.^Vr^gf. 

17.  V4  -  8  Vs. 


19.  v^54  -  9  Vl8. 

(  2a  v^8i  -  3  V243. 

21.  Vr^  -  3  /^^  V5. 


/^2  _  ^,2  Vg 


,22.  aJ- 

17.  V4-»V3.  I T^Ti 

18.  V36  +  18V5.       24.^/^^  +  (^)V3. 


25.  3  Vs. 

26.  4V3. 

27.  2  Vs. 


28.  aVa. 

29.  2cV^. 

30.  A^l. 


31.  x^V^. 


& 


a  ')9 
3N^ 


=»3-(2«  +  l)>i^ 


34. 


2c  —  3a 


125 


5        \{x-3af' 
Express  in  simplest  form  with  one  radical  sign : 

35.  \^.     38.  \/V^.  41.  VsVsVS.  44.  VVV^. 

36.  >7^.     39.  ^yVS^.   42.  >yV8.  ^5.  VV^. 

37.  VV^.     40.  VsVl.       43.  2V^2V2.      46. 


ROOTS,  RADICALS,  AND  EXPONENTS  97 

58.  Addition  and  subtraction  of  radicals.  Similar  radicals  are 
radicals  of  the  same  order  with  radicands  which  are  identical 
or  which  can  be  made  so  by  simplification. 

The  sum  or  the  difference  of  similar  radicals  can  be  ex- 
pressed as  one  term,  while  the  sum  or  difference  of  dissimilar 
radicals  can  only  be  indicated. 


EXERCISES 


Simplify  and  collect : 

1.  Vs  +  Vis. 

2.  V50  +  V98-V32. 

3.  -V^  +  ^^  -  3  ^2. 

4.  ■v^l92  -  4  ^24  +  -v^STS 

5.  ioVi-V5  +  4V¥. 

6.  3VI  +  3VJ-2V^. 

11.  </'S2  .r'  +  A^1250  X  -  -^512  X  -  v^. 

12.  W(^a-^cf  -  c  -^(a  +  c^  +  2  c  </(a  +  cf. 

•13.  ■y/(a  —  cy  +  c  ^a^  —  2  ac  +  c^  +  (a -{- c)  -^a  —  e. 


7.  a  Vx^  —  Va^x  —  5  y/a^x. 

8.  V^  +  ^-12-v^. 
Is  a   .       \3x  lax 


^-4-4^ 


+ 


N 


f^^  +  ^^ 


+  2- 


N 


a-^  +  c' 


-2. 


15.  ^24  +  v'(3  a  +  9)(a  4-  Sf  _  -v^Sl  +  a  -v^9  -  4  ^. 


16.  2  V9  a^  -  9  «2^>  -S^9ab^-9b^-{-  V(«2  _  ^,2^)  (^  _^  /^^)^ 


*«    /        7x      tT^   ,      /77F"i — ?T^7^  ,  ^H-^       36 «^ 

17.  (a  —  Z»)  -V 7  -+  V25  a^  —  25b^-\ r  -v  

^         ^    ya  —  b  a  —  by        a 


^-S6b' 


59.  Multiplication  of  real  radicals.   Radicals  of  the  same  ord^r 
are  multiplied  as  follows  : 

Example  1.    Multiply  2  Vx  —  3  Va  —  4  Vox  by  2  Vox. 

Solution :  2  V^    —  3  Va    —  4  ^s/ax 

2^  ax 

4  X  Va  —  6  a  V^  —  8  aa; 
Radicals  of  different  order  are  multiplied  as  follows : 


98         SECOND  COURSE  IN  ALGEBRA 

Example  2.    Multiply  V/i  by  "vx. 
Solution  :  V"^  =  w^  =  n*  =  -y/r^. 

■ 

Then  Vw  •  V^  =  n^  .  x^  =  V^  •  Va:^  =  Vn^. 

The  method  of  multiplying  radicals  may  be  stated  in  the 

Rule.  If  necessary,  reduce  the  radicals  to  the  same  order. 

Find  the  products  of  the  coefficients  of  the  radicals  for  the 
coefficient  of  the  radical  part  of  the  result. 

Multiply  together  the  radicands  and  write  the  product  under 
the  common  radical  sign. 

Reduce  the  result  to  its  simplest  form. 

The  preceding  rule  does  not  hold  for  the  multiplication  of  imagi- 
nary numbers ;  that  is,  for  radicals  of  even  order  in  which  the  radi- 
cands are  negative.  This  case  will  be  discussed  in  the  chapter  on 
Imaginaries. 

EXERCISES 

Perform  the  indicated  multiplications  and  simplify  the 
products : 

1.  V3.V27.  7.  Vi-Vl-V^.      IZ.-V^^-^c. 

2.  V12.  VI8.  8.  Vs.  V2.  3|-      j- 

3.  ^.^32.  9.    76. V2.  ^*-  A~o^'\x' 
4.-^20-^12.            10.  Vi2.Vi-  15.  ■v'2^3.V3x 

5.  V§.Vl-  11-  Vc-Va.  16.  (Vaj-3)l 

6.  Vl-Vl-Vl-       12.  -v^-Val  17.  (2V3a;-l)l 

18.  3  Vcc  -  3  .  V4a;-8.       20.  (  V.t  -  3  -  V4  cc  -  7)"- 

19.  (  Vi  -  Va!  -  3)'.  21-  (  Vi  -  V3^)  (4  V^). 

22.  (5V5-f9V3- V7  +  2Vi05)(V3  + V5- Vt). 

23.  1^6-2  V5Y  r(V5  +  l)(V5-f  1)T  (  V2  +  l)(9  V2- 9) 
2  16 


,^(5^. 


ROOTS,  RADICALS,  AND  EXPONENTS 


99 


24.  (c  V^  —  Vex  H-  X  Vc)  (  Vc  —  Va^). 

25.  (  Va  —  Vac  +  Vc)  (  Va  H-  Vac  +  Vc). 

26.  (  V2x  -  1  -  V5)(2  V2x  -  1  +  V45). 

Determine  which  o^  the  two  surds  is  the  greater : 

27.  vn,  Vs.  _ 

Solution  :   Vll  =  111  =  11«  =  Vll^  =  V121. 

V5=5^iU5t=  V5^-Vl25. 
Since  125  >  121,  thei  Vs  >  VlT. 

28.  Ve,  Vs.        i    30.  V3,  Ve.  32.  3  Vs,  2  Vio. 

29.  Vl9,  Vr.      .        31.  2  VS,  V89.  33.   ViS,  V64. 

Arrange  in  order  of  magnitude  : 

34.  V3,  Ve,   V^125.  35.  4  Vg,  3  ^^25,  4  V64. 

•Reduce  to  respectively  equivalent  surds  of  the  same  order : 

36.  V3,  VSs^.  38.  2 x  -V^xy,  5 x  -V^xy. 

37.  Va  4-  ^,  Va  —  5.  39.  V^,  Vcct/^,      ' 

Square : 

40.  V3. 

41.  2V4. 
Cube: 

46.  V3. 

47.  3V5. 
Simplify  : 


xy. 


42.  V5  -  Vs. 


43.  4  \/3  -  Vs. 

48.  V5-V3.  

49.  V3  -  Vs.  51.  {^1  -  V2)'. 


44.  V4  -  4  Vs. 

45.  Ve  -  3  V2. 

50.  3  V2  -  2  V3. 


53 


56 


■^RJ 


(e^  -  2  e-*-)  (e^  -  2  c"^')  +  e'^^  +  c'^^  -  6. 


100  SECOND  COURSE  IN  ALGEBRA 

60.  Division  of  radicals.  Division  of  radicals  is  usually  an 
indirect  process  performed  by  means  of  a  rationalizing  factor 
for  the  divisor. 

One  radical  expression  is  a  rationalizing  factor  for  another  if 
the  product  of  the  two  is  rational. 

Thus  a  Vti  and  -\n  are  rationalizing  factors  for  each  other. 
A  like  relation  holds  between  a  V/i  and  V^. 

An  important  pair  of  radicals  is  Va  -|-  V&  and  Va  —  Vft. 
Two  such  binomials  are  called  conjugate  radicals  and  either  is 
the  rationalizing  factor  for  the  other. 

Division  of*  one  radical  by  another  may  often  be  performed 
as  follows : 

Example :  Divide  6  V5  by  3  VS. 

Solution :  6V5^3V3  =  2V|  =  |  ViS. 

Direct  division  of  radical  expressions  in  which  the  divisor  is 
a  polynomial  is  very,  difficult.    In  such  cases  we  use  the 

Rule.  Write  the  dividend  over  the  divisor  in  the  form  of  a 
fraction.  Then  multljjlij  the  numerator  and  denominator  of  the 
fraction  by  a  rationalizing  factor  for  the  denominator  and  sim- 
plify the  resulting  fraction. 

This  rule  applies  in  all  cases,  while  the  rule  for  direct  division 
fails  when  dividing  a  real  radical  by  a  radical  of  even  order  whose 
radicand  is  negative. 

Every  irrational  algebraic  expression  containing  nothing  more 
complicated  than  rational  numbers  and  radicals  has  a  rationalizing 
factor.  To  find  this  factor  for  any  given  irrational  expression  is  a 
problem  which  requires  considerable  algebraic  training.  At  the 
present  time  it  is  wholly  beyond  the  student  to  find  the  rationalizing 
factor  of  even  so  simple  an  expression  as  the  denominator  of  the 

.      . .  A^  +  Va 

traction — — -— -—  •    The  approximate  value  of  such  a  fraction 

can  be  obtained,  however,  by  dividing  the  sum  of  the  approximate 
values  of  the  terms  in  the  numerator  by  the  sum  of  the  approximate 
values  of  the  terms  in  the  denominator.  (The  table  on  page  262  may 
be  used  to  advantage  in  work  of  this  character.) 


EOOTS,  RADICALS,  AlS^D'EXj^tENTS  101 

Pind  a  simple  rationalizing  factor  for : 

1.  3  Vt.  4.  Vs.  7.  Ve  -  Vn. 

2.  5  y/l.  5.  Vl6.  8.  3  Vz  -  2  VlS. 

'  3.  7^8.  6.  V5  -  7.  9.  Vs^  -  V2^. 

10.  Vic-c- V^.  11.  V2  +  V3-V5. 

Perform  the  indicated  division  and  simplify  results  : 

12.  V12--V3.  14.  8^4V3.  16.  24 -f- 3  Vs. 

13.  V8^V24.  15.  8--2V2.  17.  a.H-cV^. 

18.  (Vl^- Vl8)--2V3. 

19.  (12  -  3  Ve  -  4  V24)h-  3  V2. 

20.  V6-=-V2. 

Hint.  V^.V2  =  ^  =  ^^  =  ^^^  =  etc. 
V2       V2V4  2 

21.  VS  -  V2.  22.  V32  ^  V2.        ^  23.  v'i  -  V^- 

24.  3^(2-V3). 
HixT.   3^(2-^)^-^^^         3(2+V3)         ^^^^^ 
2  -  V3       (2  -  V3)  (2  +  V3) 

25.  4-(V2-l).  27.  V7-f-V2-V3. 

26.  V3^(V2H-V3).  28.  V^^Vl^Vs^Vl. 
29.  (V7  + V5)^(2V7- V5)^(19-3V35). 

30.  Find  to  four  decimals  the  numerical  value  of  the  results 
in  Exercises  (a)  24,  (b)  26,  and  (c)  27. 

31.  In  Exercise  26  divide  the  numerical  value  of  the  numer- 
ator by  the  numerical  value  of  the  denominator,  each  having 
been  obtained  to  four  decimals.  Compare  the  quotient  with  the 
result  obtained  for  that  fraction  in  Exercise  30. 

32.  What  conclusion  can  be  drawn  from  Exercises  30  and 
31  regarding  the  rationalization  of  the  denominator  of  a  frac- 
tion before  finding  its  numerical  value  ? 


102  SEtlOiND  COURSE  IN  ALGEBRA 

Changed  to  respect  J  Vefy'  equivalent  fractions  having  rational 
denominators : 

J-^    V5  +  V2  V^  +  V^  -_     -y/x-S  +  VS 

««>  ~~j= ;;=•  35.  — 7= 7=-         37.  — .  7=- 

-     V5-V2  Va-Vb  Vic  -  3  -  V3 


.0'  3V5-2V7  '     Vx  +  Vc  ^2-V2 

Perform  the  indicated  division  : 
39.  (VlO-V5)--(Vi04-V5).      40.  {x  -  Vc)  ^  {x  -  S  V^). 

41.  (Va  +  c  —  V^)-i-(Va4-c  +  Vi). 

42.  (V3  +  V2)h-(2-V3  +  V2). 

43.  (V5-V7)-f-(V5-f  V7-V2). 

44.  Is  there  any  real  distinction  between  the  direction  before 
Exercise  33  and  that  before  Exercise  39  ? 

45.  Does  3  -  Vt  satisfy  x'-6x-\-2  =  0? 

46.  Does  ^~r,    ^  satisfy  2x^ -75x +  161  =  0? 

47.  Does  ^(5  ±  VI09)  satisfy  3aj2_5^_7^Q^ 

61.  Square  root  of  surd  expressions.  The  square  of  a  binomial  is 
usually  a  trinomial.  However,  the  result  of  squaring  a  binomial  of 
the  form  Va  +  Vft  is  a  binomial,  if  a  and  b  are  rational  numbers. 
Thus  (V7-V^)2^7-2V21  +  3  =10-2  V2I.  Here  in  10 - 2  V2I, 
10  is  the  sum  of  7  and  3,  and  21  is  the  product  of  7  and  3.  These 
relations,  and  the  fact  that  the  coefficient  of  the  radical  V2I  is  2, 
enable  us  to  find  the  square  root  of  many  expressions  of  the  form 
a  ±2  -y/b  by  writing  each  in  the  form  of  a:  ±  2  -y/x^  +  y  and  then 
taking  the  square  root  of  the  trinomial  square  as  follows : 

Example :  Extract  the  square  root  of  9  —  V56. 
Solution :  9  -  V56  =  9-2  Vl4. 

We  now  find  two  numbers  whose  sum  is  9  and  whose  product 
is  14.    These  are  7  and  2. 

Therefore  9-2  Vl4  =  2-2  Vli  +  7  =  ( V2  -  V7)2. 
Hence  the  square  roots  of  9  —  V56  are  ±  ( V2  —  V7). 


ROOTS,  RADICALS,  AND  EXPONENTS  103 

EXERCISES 

Find  the  positive  square  roots  in  Exercises  1-12 : 

1.  6- 2  Vs.  3.  13  +  V48.  5.  11  -  4  Vt. 

2.  7  +  2V1O.  4.  8-V60.  6.  I7  +  I2V2. 

7-11-3V8.  10.  1|^_V3^. 

8.  65x-200V3^. 


11.  2ic  +  2V^2-49. 
9.  126a-10aV5.  ^^    ^  +  V^^^T. 

13.  V9  +  3  VS  =(  V?  +  V?)l  

14.  Vl5  -  5  Vs  =  ?  15.  Vc^  4-  Va2-4^>2  =  ? 


16.  Vm"^  -\-vi  +  2n  +  2m  Vm+~2 


71 


? 


Note.  In  the  writings  of  one  of  the  later  Hindu  mathematicians 
(about  1150  A.D.)  we  find  a  method  of  extracting  the  square  root  of 
surds,  which  is  practically  the  same  as  that  given  in  the  text.  In  fact, 
the  formula  for  the  operation  is  given,  apart  from  the  modern  symbols, 

as  follows  :  Va  +  v^  =  v  «  +  &  +  2  -yUih.    The  study  of  expressions 

of  the  type  v  V«±V6  had  been  carried  to  a  most  remarkable 
degree  of  accuracy  by  the  Greek,  Euclid.  His  researches  on  this 
subject,  if  original  with  him,  place  him  among  the  keenest  mathe- 
maticians of  all  time  ;  but  his  work  and  all  of  his  results  are  expressed 
in  geometrical  language,  which  is  very  far  removed  from  our  algebraic 
symbolism,  and  for  that  reason  is  little  read  now. 

62.  Factors  involving  radicals.  In  the  chapter  on  Factoring  it  was 
definitely  stated  that  (except  in  §  17)  factors  involving  radicals  would 
not  then  be  considered.  This  limitation  on  the  character  of  a  factor 
is  no  longer  necessary.  Consequently  many  expressions  which  pre- 
viously have  been  regarded  as  prime  may  now  be  thought  of  as  fac- 
torable ;  thus 

3a:2-l  =  (a:V3  +  l)(x  V3-I)  and  42-2- 5=  (2:c  +  V5)(2x- Vs). 

It  is  not  usual  to  allow  the  variable  in  an  expression  to  occur  under 
a  radical  sign  in  .the  factors.  Hence,  if  a:  is  a  variable,  the  trinomial 
a;2  +  a:  + 1  is  not  regarded  as  factorable  into  {x  +  -\/x  +l)(x  —  -y/x  -M), 
though  (a;  4-  V^  -1- 1)  (x  -  V^  +  1)  =  a;2  +  a:  +  1. 

Therefore  in  this  extension  of  our  notion  of  a  factor  it  must 
be  clearly  understood  that  the  use  of  radicals  is  limited  to  the 


104 


SECOND  COURSE  IX  ALGEBRA 


coefficients  in  the  terms  of  the  factors.  Such  a  concejition  of  a  fac- 
tor is  a  necessity  for  certain  work  in  advanced  algebra  and  geometry, 
and  is  very  desirable  in  solving  equations  by  factoring. 

To  restrict  the  use  of  radicals  in  the  way  just  indicated  is  neces- 
sary for  the  sake  of  definiteness.  Otherwise  it  M'ould  be  impossible 
to  obey  a  direction  to  factor  even  so  simple  an  expression  as  x^  —  y-\ 
for  if  the  variable  is  allowed  under  a  radical  sign  in  a  factor,  x^  —  ip- 
has  countless  factors. 

Thus  x'-y''  =  {j^^y){x-y) 

=  {x.->r  y){^-\-^j){Vx-^/y) 

=  (x  +  y)  (Vx  -t-  Vy)  (-y/x  -f  Vy)  (  V^  -  V^)  =  etc. 


Factor : 

1.  x""  -  11. 

2.  3ic=-16. 


EXERCISES 

3.  ic«  +  2. 

4.  x""  - 12. 


5.  3.T«-27. 

6.  5  x» -1-125. 


Find  the  algebraic  sum  of : 

7.1^  +  — 2 

a-b      V«  -f  V^ 

Solve  by  factoring  and  check : 


r^  -f-  c^ 


ViC  —  Vc 


9.  ir^  -  5  =  0. 

11.  ^^-f  144  =  26a^l 

10.  2x^-3  =  0. 

12.  4:X^-{-c  =  x^-\-4:Cx\ 

MISCELLANEOUS  EXERCISES 
Solve  for  71 : 

1.  €1^.0"  =  a\ 

6.  42.22  =  2".          11.  22.2"  =  32. 

2.  2^-22  =  2". 

7    4«.2='  =  2".          12.  82.4«  =  2''. 

3.  2*.2«  =  2". 

8.  42.2^  =  22".         13.  8^.42  =  2". 

4.  2^  •  2"  =  2^\ 

9.  3^ •9'^  =  3".          14.  9^.  272  =  3", 

5.  32-2^  =  9. 2«. 

10.  92.3"  =  3^          15.  27" •92  =  3^'^ 

16.  8«.42«  =  2". 

17.  3«.9''  =  81l 

18.  9».3«=27«. 

n 

20.  81 .  27"  =  (9")2. 

19.     22«  +  2.4n  +  2  _ 

82".               22.  2«"  +  ^.4«"  +  «  =  (8")". 

ROOTS,  RADICALS,  AND  EXPONENTS 


105 


Solve  for  x : 

23.  x3  =  8. 

24.  x'^  =  6. 
25.^-1  =  256. 


26.  x^  =-  343. 

27.  \x-i  =  2. 
28>  (a;-2)-*=49. 


29.  (ax^)-«  =  27. 
-yj      -^16 


f.  36.  {x-'^^^Vi)\ 

42.  vV«: 

43.  V  ;5/^. 

8/    4,— 

44.  V  Vxl 


Express  in  simplest  form  with  positive  exponents : 

31.  (^y:^277«)-l     33.^^16^.  35.  (:^iV^ 

32.  (A/l6"aV)"'.      34.  {x-Vli 

37.  u-^v^^^ri 

38.  [(-^/S^n'. 

39.  {^  </Wa')\ 

41.  [Jx'i  \ax~i  y/x^. 
Simplify: 

48.  (a«-i).(a^-")^(a2«  +  5)-i, 

n  +  2  ■  _J_ 

49.  a"  +  i  -T-a'^  +  i. 

50.  a"^^  H-  a"+^. 


45.  V2a^^. 
Voce 


46 


55. 


a'  •  16^( 


51.  a»"-"^  ^  a"- 

1 

52.  {{a^  +  y-^y 


«   ^  +  — 

56.  [(x  -  ^)2]^  (if  ;r  ^  2/). 

57.  (a;^"')^. 
,  58. 


•2s' 


2l.-^ 


53.  [((a;-'")")-^-[((a;-«)"7].59.  [(2xy.(2x)^(2x)-']-i 

54.  (^y^-^^x)^A  60.  [32-t.9^]-i 

4n  +  l  gn  +  l 


T  +  3. 


2"(4''-i)«      (4«  +  i)'' 
\125  7/-V     \A~V     \icyy'\7>^V     V 


.6\i 


106  SECOND  COURSE  IN  ALGEBRA 

63.  [(5a)2».(6a.)''-«»]l  66.  -^^^ -{- 3(x-yK 

64.  (J-a-^^)^.  3a-^c-«    21x-''i/-'^ 

(2- 3)" -3""*     *         1^"^' 
m^Ti      m''r''  +  ^  __  xy^z -\/xy  y/l 


70.  — — T  -ZT^:^-  72 


Find  the  square  roots  of : 

73.  £c^  +  4  o^V^  +  9  —  4  x^y^  +  6  a;^  -  12  x-^^/i 

Find  the  indicated  roots  of  : 

75.  [(e^  -  e-'^y  +  4]'.  76.  [((f  +  2-ie-^)2-  2]'^. 

77.  (e^^  +  e-2^  +  2  +  4  e^  -  4  e"^)*. 

78.  (x-^  +  llx-^  +  16^2  _  5^-4  _  24)^. 

79.  /--^  +  6a;-i-4ic-^  +  lV. 

on     /—     -      ,    -^-'^*    ,    16    ,  e/e        16  ^-'^^        Q>a-%^\^ 

80.  — 7^^ 1 tt; 1-  773  +  a-%^ 


/4a-«^>«       9  a 
M^      9      "^      J 


16       '  25   '  15 

Simplify : 

x^-a-x""  —  a^x^  (x^  —  l)ax''-'^  —  x''-2x 

81.  ^- 


a 


82.  — 


(xy 


x'^  +  20 


83. 

(x^-iy 

x^ax'' 

x" 

x'-l 

l)2a^ 

84. 

(x^r 

x^  +  1 

85. 


ROOTS,  RADICALS,  AND  EXPONENTS 


lOT 


x-\-2x-^)  -  jx"  +  3)  (-  5x-^) 

^ {x-'^f 

0^-2  +  3 


87. 


2  e"*  +  1     . 


88. 


-v^  +  1 


Vaa;  —  x^ -h  —  hx {ax  —  x^)    ^  •  (a  —  2x) 


89. 


90. 


91. 


( -\/ax  —  x^Y 

hx  H-  '^ ax  —  cc^ 

2  7l^2n-l_^2«(^^2«. 

-l)"*(^a;' 

(Vx2" 

-1) 

"-^) 

(V^2n_iy-^ 

•5-5cc.-|-(cc2-10 

a;)-*(2a;- 

Vic2_ 

-lOcc 

-10) 

(Vo^^-lOa.)^ 

92.  Show  that 


93.  Show  that 


hx^ix^-x^xy 


Vm  +  V^       Vm  —  7Z.       Vm  —  ti 


V'^        "".J/^ 


V'-^^       •v^-'"--v^ 


108  SECOND  COURSE  IN  ALGEBRA 

PROBLEMS 

(Obtain  answers  in  simplest  radical  form.) 

1.  The  side  of  an  equilateral  triangle  is  12;  find  the 
altitude. 

2.  The  side  of  an  equilateral  triangle  is  s;  find  the  alti- 
tude and  the  area. 

3.  The  altitude  of  an  equilateral  triangle  is  20;  find  one 
side  and  the  area. 

4.  Find  the  side  of  an  equilateral  triangle  whose  altitude 
is  a. 

5.  Find  the  altitude  on  the  shortest  side  of  the  triangle 
whose  sides  are  9,  10,  and  17.    Find  the  area  of  the  triangle. 

6.  Find  the  altitude  on  the  longest  side  of  the  triangle 
whose  sides  are  10,  12,  and  16. 

D  In  the  adjacent  regular  hexagon 

/  \  AB  =  BC  =  CD,  etc.    0  is  the  cen- 

/  \  ter  and  OK  is  the  altitude  of  the 

/  \  hexagon. 

)C'  7.  Find  the  altitude  and  the  area 

/  of  a  regular  hexagon  (a)  whose  side 

\      /  is  15 ;  (b)  whose  side  is  s. 

\/  Fact  from  Geometry.  A  regular  hexa- 

-^        J^         -B  gon   may   be    divided    into    six  equal 

equilateral  triangles  by  lines  from  its  center  to  the  vertices. 

8.  Find  the  side  and  the  area  of  a  regular  hexagon  (a)  whose 
altitude  is  25 ;  (h)  whose  altitude  is  h. 

9.  The  base  of  a  pyramid  is  a  square,  each  side  of  which 
is  10  feet.  The  other  four  edges  are  each  20  feet.  Find  the 
altitude  and  the  volume  of  the  pyramid. 

Fact  from  Geometry.    The  volume  of  a  pyramid  or  cone  is  — — » 
where  a  is  the  altitude  and  h  is  the  area  of  the  base. 

10.  The  side  of  an  equilateral  triangle  is  18.    Find  the  two 
parts  into  which  each  altitude  is  divided  by  the  other  altitudes. 


ROOTS,  RADICALS,  AND  EXPONENTS 


109 


Fact  from  Geometry.  The  altitudes  of  an  equilateral  triangle  inter- 
sect at  a  point  which  divides  each  altitude  into  two  parts  whose  ratio 
is  2  to  1. 

The  altitude  of  a  regular  tetrahedron  (DK  in  the  adjacent 
figure)  meets   the  base  at  the 
point  where  the  altitudes  of  the 
base  intersect. 

11.  A  BCD  is  a  regular  tetra- 
hedron. If  each  edge  is  12,  find. 
CJi,  CK,  and  lastly  the  alti- 
tude DK. 

Fact  from  Geometry.  A  regular 
tetrahedron  is  a  pyramid  whose 
four  sides  are  equal  equilateral 
triangles. 

12.  Find  the  altitude  and  vol- 
ume of  a  regular  tetrahedron 
whose  edge  is  15. 

13.  Show  that  the  altitude  and  the  volmne  of  a  regular  tet- 

rahedron  whose  edge  is  e  are  respectively  -  V 6  and  —  V 2. 


CHAPTER  VII 

GRAPHICAL  SOLUTION  OF  EQUATIONS  IN  ONE  UNKNOWN 

63.  Functions.  An  algebraic  expression  involving  one  or 
more  letters  is  a  function  of  the  letter  or  letters  involved. 

The  letters  of  a  function  are  usually  referred  to  as  variables. 

A  function  is  called  linear,  quadratic,  or  cubic  according  as  its 
degree  with  respect  to  the  variable  or  variables  is  first,  second, 
or  third  resj)ectively. 

Examples  of  the  functions  just  named  are  respectively 
4ic-7,  2x''-5x  +18,  z^-\-8z^-2z-6. 

After  a  function  of  any  variable,  say  x,  has  once  been  given, 
it  is  convenient  and  usual  to  refer  to  it  later  in  the  same  dis- 
cussion by  the  symbol  f(x),  which  is  read  the  function  of  x,  or 
more  briefly  fofx. 

64.  Graph  of  a  function.  A  graph  always  shows  a  relation 
between  (at  least)  two  variables.  The  graph  of  a  function  of 
one  variable  is  a  curve  showing  the  value  of  the  function  for 
any  real  value  of  the  variable.  This  means  that  one  axis  must 
be  the  as'-axis  and  the  other  the  function  axis,  or  J'-axis.  The 
method  of  constructing  the  graph  of  a  function  of  x  is  the 
same  for  a  linear,  a  quadratic  (see  "First  Course  in  Algebra," 
pages  259-266),  and  a  cubic  function  in  one  variable. 

65 .  Graph  of  a  cubic  function.  To  graph  the  function  x^—  5  ic  +  3, 
first  prepare  a  table  of  values  as  follows : 


When                   X  = 

-1 

-3 

-2 

-1 

0 

1 

2 

n 

3 

f(x),x^-ox  +  n  = 

-41 

-9 

5 

7 

3 

-1 

1 

H 

15 

110 


GRAPHICAL  SOLUTION  OF  EQUATIONS 


111 


Plotting  the  points  corresponding  to  the  numbers  in  the  table 
(except  the  first  and  last),  we  obtain  A  (-  3,  -  9),  B,  C,  D,  E,  G, 
and  //,  in  the  order  named.    The  curve  crosses  the  x-axis  three  times  : 


F 

/ 

/ 

c 

/ 

/ 

\ 

/ 

S 

/ 

, 

/" 

\ 

/ 

/ 

)^ 

/ 

/ 

^ 

/ 

/ 

\ 

4- 

y' 

/ 

'   \ 

y 

-' 

I 

- 

1 

0 

\ 

/ 

"' 

E 

-2 

1 A 

f' 

once  betwfsen  1  and  2 ;  again  between  0  and  1 ;  and  a  third  time 
between  —  2  and  —  3.  At  the  points  of  crossing /(x)  is  zero.  There- 
fore the  values  of  x  at  these  points  are  the  roots  oi  x^  —  5  x  +  3  =  0. 
These  are  approximately  1..8,  .6,  and  —  2.5. 


I 


EXERCISES 

(Exercises  12-16  refer  to  the  preceding  graph.) 

1.  Construct  the  graph  oif(x)  =  Sx  —  9. 

2.  Does  the  ic-coordinate  of  the  point  where  the  line  crosses 
the  a;-axis  satisfy  the  equation  3  a?  —  9  =  0  ?    Why  ? 

3.  What  is  the  graph  of  any  linear  function  ot  x? 

4.  Construct  the  graph  of  f(x)  =  2  x^  —  x  —  6. 

5.  Do  the  a!;-coordinates  of  the  points  where  the  curve  crosses 
the  a;-a2?:is  satisfy  the  equation  2x^  —  x  —  G  =  0?    Why  ? 


112 


SECOND  COURSE  IN  ALGEBRA 


6.  What  kind  of  a  line  do  you  expect  the  graph  of  any 
quadratic  function  in  one  variable  to  be  ? 

7.  State  a  rule  for  the  graphical  solution  of  a  linear  or  a 
quadratic  equation  in  one  variable. 

8.  What  is  the  effect  on  the  graph  of  a  quadratic  function 
of  07,  if  a  positive  number  is  added  to  the  constant  term  ? 

9.  What  change  occurs  in  the  roots  of  a  quadratic  equation 
in  X,  if  a  positive  number  is  added  to  its  constant  term  ? 

10.  When  does  the  graphical  solution  of  a  quadratic  equa- 
tion give  but  one  real  root? 

11.  When  does  the  graphical  solution  of  a  quadratic  equa- 
tion fail  to  give  the  roots  of  the  equation  ? 

12.  If  the  function  x'  —  5  a:  -}-  3  be  set  equal  to  4,  can  the 
roots  of  the  equation  thus  formed  be  read  from  the  graph  ?  If 
so,  read  them. 

13.  Set  a;'  —  5  aj  +  3  equal  to  —  1.3  (approximately)  and  read 
the  roots  of  the  resulting  equaticfn  from  the  graph.    Explain. 

14.  Set  the  function  cc^  —  5  ic  +  3  equal  to  —  4  and  read  the 
roots  of  the  resulting  equation  from  the  graph.    Explain. 

15.  Set  x^  —  5  aj  +  3  equal  to  8  and  read  the  roots  of  the 
resulting  equation  from  the  graph.    Explain. 

16.  Set  f(x)  equal  to  9  and  read  the  roots  of  the  resulting 
equation  from  the  graph.    Explain. 

17.  (a)  Is  a  rational  function  always  integral  ?  (h)  Is  an 
integral  function  always  rational?  (c)  Write  an  example  of 
each. 

66.  Imaginary  roots.  To  make  clearer  the  point  in  Exercises 
14-16  preceding,  we  shall  graph  the   function  ic^  —  2  .t  —  4. 


When                 X  = 

-3 

-^ 

-2 

-1 

0 

1 

2 

^ 

o 

f(x),3^-2x-i  = 

-25 

-14| 

-8 

-3 

-4 

-5 

0 

H 

17 

GRAPHICAL  SOLUTION  OF  EQUATIONS  113 


The  point  A  corresponds  to  -  2\,  -  14f  •  The  points  correspond- 
ing to  the  next  six  pairs  of  numbers  given  are  B,  C,  D,  E,  G,  and 
H  in  the  adjacent  figure.  The  curve 
through  these  points  crosses  the  a;-axis 
but  once.  This  shows  that  the  equation 
has  but  one  real  root,  and  that  the 
value  of  this  root  is  2.  Since  the  num- 
ber of  roots  of  a  rational  integral  equa- 
tion is  the  same  as  the  number  which 
indicates  its  degree,  we  conclude  that 
the  other  two  roots  are  imaginary. 

Note.  It  required  the  genius  of  Sir 
Isaac  Newton  first  to  observe  from  the 
graph  of  a  function  that  two  of  its  roots 
become  imaginary  simultaneously.  He 
also  saw  that  an  equation  with  two  of 
its  roots  equal  to  each  other  is,  in  a 
certain  sense,  the  limiting  case  between 
equations  in  which  the  corresponding 
roots  appear  as  two  real  and  distinct 
roots,  and  those  in  which  they  appear 
as  imaginarv  roots. 


F 

H       T 

jr 

i 

5-     i 

4                   4 

1     i 

2                  t 

I                 t 

i__G--X 

0    1^"-^ 

J 

7 

Zc"N 

D     J 

_i 

>^2 

7 

■5       E 

p 

-4^ 

1 

1 

1 

t 

A 

1— 

67.  Graphical  solution  of  an  equation  in  one  unknown.    If  the 

student  has  grasped  the  meaning  of  the  preceding  graphical 
work,  he  will  see  the  correctness  of  the  following  rule  for  solv- 
ing graphically  any  equation  in  one  unknowil. 

Rule.  An  equation  in  one  unknown  whose  second  ineTnhev  is 
zero  is  solved  for  real  roots  by  graphing  the  function  in  the  first 
"tnonher  and  then  obtaining  the  value  of  x  for  the  points  tuhere 
the  curve  crosses  the  x-axis. 


68.  More  accurate  graphical  solutions.  By  drawing  the  entire 
graph  to  a  larger  scale  the'  student  can  obtain  more  accurately 
the  values  of  the  roots.  If  still  more  exact  results  are  desired, 
he  may  proceed  somewhat  as  follows  : 

The  graph  on  page  111  shows  that  there  is  a  root  greater  than  .6 
and  less  than  .7.  If  we  now  construct  on  a  large  scale  that  portion 
of  the  curve  between  D  and  E  (page  111)  which  is  just  above  and 


114 


SECOND  COURSE  IN  ALGEBRA 


just  below  the  r-axis,  we  shall  get  a  more  precise  value  for  the  root. 
Substituting  .0  and  .7  in  x^  —  5  x  -\-  3,  we  obtain  the  following  table  : 


When                  X  = 

.G 

.7 

f(x),x^-5x-^^  = 

.216 

-  .157 

I 


^ 


Between  x  =  .6  and  x  =  .7  the  func- 
tion changes  from  +  to  — .  Hence  we 
are  certain  that  the  graph  crosses  the 
X-axis  between  these  points.  We  now 
choose  a  much  larger  scale  than  the 
one  used  on  page  111.  This  is  indi- 
cated by  the  numbers  on  the  x-axis. 
The  scale  is  too  large  to  show  the 
y-axis  in  the  figure,  so  the  scale  for 
y  is  indicated  on  the  line  AB.  The 
point  K  corresponds  to  .6,  .216,  and 
the  point  L  to  .7,  —  .157.  Since  K  and 
L  are  comparatively  close  together, 
the  portion  of  the  graph  between  them  is  nearly  a  straight  line. 
Drawing  the  straight  line  KL,  it  is  seen  to  cross  the  x-axis  between 
.64  and  .66,  or  about  .658.  By  an  algebraic  method  of  solution  it 
can  be  shown  that  the  root,  correct  to  three  decimals,  is  .656.  Here 
the  graphical  method  gives  the  result  to  within  -^^  of  one  per  cent 
of  the  true  value. 

EXERCISES 

Solve  graphically : 

,  (Obtain  roots  in  Exercises  2  and  6  correct  to  two  decimals.) 


\ 


i 


x^^U  =  Sx. 


^7.  x'-10x''^16  =  0. 
Sf  x'-4:X^-\-12  =  0. 
By  reference  to  the  curve  obtained  in  Exercise  6  solve  : 
"^9.   (a)  x^-4:X  =  -5.    (b)  x^-10  =  4.x.    (c)x^-4.x-2 


2.  ic^-3ic  +  4  =  0. 

3.  x^-}-x  =  4:. 

4.  x''-2x^-5x-{-6  =  0. 


0. 


69.  Critical  values  of  the  variable.    There  are  many  practical 
problems  involving  two  variables  in  which  it  is  necessary  to 


GRAPHICAL  SOLUTION  OF  EQUATIONS  115 

determine  that  value  of  one  variable  for  which  the  other  has 
the  greatest  (or  least)  possible  value.  A  great  number  of  these 
l^roblems  can  be  solved  by  means  of  a  graph.  The  method 
of  solution  can  be  made  clear  by  reference  to  the  graph  of 
§  65.  There  /(j?)  =  x^  —  5  x  -\-  S.  Suppose  we  wish  to  know 
the  value  of  x  which  gives  x^  —  5x  -{-  3  the  greatest  possible 
value.  Near  C  occurs  a  high  point,  —  a  turning  point  of  the 
curve, —  and  there  a?  =  —  |  approximately,  and/(x)  =  7.3.  This 
value  of  X  gives  to  f(x)  a  greater  value  than  does  any  other 
value  of  X  between  —  2.5  and  -f  2.6.  It  is  true  that  on  the 
portion  GH  above  H  greater  values  of /(ic)  than  7.3  occur.  But 
in  practical  problems  similar  to  those  of  the  next  list  it  will 
be  found  that  some  condition  of  the  problem  will  rule  out  of 
consideration  any  value  of  x  which  does  not  correspond  to 
the  turning  point  of  the  curve  such  as  that  which  occurs  near 
C  or  near  E. 

PROBLEMS 

1.  A  manufacturer  has  in  stock  a  quantity  of  strawboard  8 
inches  by  15  inches,  out  of  which  he  desires  to  make  open-top 
boxes  by  cutting  equal  squares  out  of  each  corner  and  folding 
up  so  as  to  make  sides  and  ends.  What  must  the  side  of  the 
square  be  so  as  to  make  a  box  of  the  greatest  possible  volmne  ? 

Hint.  Let  x  equal  the  side  of  the  square.  Then  the  dimen- 
sions of  the  box  in  inches  are  15  —  2  a:,  8  —  2  a:,  and  x.  Hence  the 
volume  V=  4  a:^  —  46  a:^  +  120  x  in  cubic  inches.  Construct  the  graph 
of  V=  4  a-3  -  46  a;  2  +  120  x  (or  that  of  F/4  =  x^-  11 1  x^  +  30  x,  which 
deals  with  smaller  numbers) ;  then  an  inspection  of  the  turning  points 
will  give  the  required  value  of  x. 

2.  Eeferring  to  the  graph  of  Exercise  1 :  (a)  What  value 
has  the  function  4  x^  —  46  a;^  +  120  x  when  x  =  If  ?  (b)  What 
other  value  of  x  gives  the  function  the  same  value  ?  (c)  What 
values  of  x  give  the  function  greater  values  than  this  ?  (d) 
What  condition  of  the  problem  rules  out  these  values  as  sides 
of  the  square  ? 


116  SECOND  COURSE  IN  ALGEBRA 

Biographical  Note.  The  notion  of  a  function  is  one  of  the  most  fun- 
damental ideas  in  modern  mathematics.  Only  the  simplest  examples  are 
given  in  this  book,  but  many  others  involving  expressions  of  the  utmost 
complexity  have  been  studied  by  mathematicians  for  many  years.  An 
important  reason  for  the  study  of  functions  is  found  in  the  fact  that  all 
kinds  of  facts  and  principles  which  we  meet  in  the  study  of  nature  can 
be  expressed  symbolically  by  means  of  functions,  and  the  discovery  of 
the  properties  of  such  functions  helps  us  to  understand  the  meaning  of 
the  facts.  A  complete  understanding  of  the  laws  of  falling  bodies,  light, 
electricity,  or  sound  could  never  be  reached  without  the  study  of  the 
mathematical  functions  which  these  phenomena  suggest. 

One  of  the  foremost  living  scholars  who  has  discovered  many  prop- 
erties of  the  most  complicated  functions  is  Professor  Felix  Klein  of 
Gottingen,  Germany.  Since  the  time  of  Gauss,  who  was  also  a  professor 
at  Gottingen,  the  university  there  has  been  one  of  the  leading  insti- 
tutions of  the  world  in  the  study  of  mathematics.  It  is  interesting  to 
know  that  Klein's  great  achievements  in  advanced  mathematics  have 
not  caused  him  to  forget  the  difficulties  which  surround  the  beginner 
in  the  first  years  of  his  study,  but  that  he  has  had  wide  influence  in 
improving  mathematical  instruction  in  the  schools  not  only  of  Germany 
but  of  other  countries  as  well. 

3.  A  piece  of  tin  is  8  inches  by  12  inches.  From  each  cor- 
ner a  square  whose  side  is  x  inches  is  cut  out.  The  sides  are 
then  turned  up  and  an  open  box  is  formed,  which  has  the 
greatest  possible  volume.    Find  graphically  this  value  of  x. 

4.  What  value  of  x  gives  ic^  —  4  £c  -f-  6  the  least  possible  value  ? 

5.  An  open  metal  tank  having  a  volume  of  4  cubic  yards  has 
vertical  sides  and  a  square  base.  Determine  the  side  of  the 
base  and  the  altitude  of  the  tank  if  the  inside  surface  is  the 
least  possible. 

Hint.  Let  x  equal  the  side  of  the  base  in  yards  and  d  the  altitude 
in  yards.  Then  the  volume  of  the  tank,  4  cubic  yards,  equals  dx^,  and 
the  surface  equals  x^  -Y  ^  dx  in  square  yards.    From  these  two  state- 

-I  /»  -t  n 

ments  we  obtain  surface  S  =  x'^  ■\ Plot  the  function  x^  ■\ and 

X  x 

the  required  value  of  x  will  be  apparent. 

6.  An  open  metal  tank  having  a  volume  of  4  cubic  yards  is 
in  the  form  of  a  cylinder  with  a  circular  base.  Determine  the 
radius  of  the  base  and  the  altitude  so  that  the  inside  surface 
will  be  the  least  possible. 


FELIX  KLEIN 


GKAPIIICAL  SOLUTION  OF  EQUATIONS  117 

7.  The  perimeter  of  a  rectangle  is  20  rods.  Find  the  length 
and  the  width  if  the  area  is  the  greatest  possible. 

8.  A  boatman  6  miles  from  the  nearest  point  of  the  beach 
(which  is  straight)  wishes  to  reach  in  the  shortest  possible 
time  a  place  8  miles  from  that  point  along  the  shore.  He  can 
row  4  miles  per  hour  and  jog-trot  6  miles  per  hour.  Determine 
where  he  must  land. 

Hint.  Draw  a  right  triangle  ABC,  AC  being  the  shore  line,  B 
the  boat,  and  A  the  point  on  shore  nearest  B.  Let  K  on  AC  be 
the  point  at  which  he  lands,  and  let  KA  in  miles  be  x.  Then 
JiK  =  -\/x^  +  36  and  CK  =  8  —  x.    In  hours  the  time  required  to  go 

from  B  to  A'  is : ,  and  that  from  K  to  C  is Therefore 

4  6 

8  —  X       s/x'^  +  36 

the  total  time  equals 1 Plot  this  function  and  the 

6  4 

required  value  of  x  will  be  apparent. 


CHAPTER  VIII 

QUADRATIC  EQUATIONS 

70.  Solution  by  completing  the  square.  Any  quadratic  equa- 
tion in  one  unknown  of  the  general  type  ax^  -\- bx -\- c  =  0  can 
be  solved  as  follows  : 

Example :  Solve  3  x^  -  7  ic  -  20  =  0.  (1) 

Solution :  Transposing,  3  a:^  -  7  a;  =  20.  (2) 

7  X      20 

Dividing  (2)  by  the  coefficient  of  x^,  x"^ —  =  — .  ,  (3) 

o         o 

Adding  (—  ^)2  to  each  member  of  (3), 

^'-Ix  +  i-  ly  =  -V- J-  li  =  ^%'-.  (4) 

Then  •         (^-|)^  =  (-y-)^.  C-^) 

Extracting  the  square  root  of  each  member  of  (5), 

Whence  x  =  l±  V-  =  4  or  -  |. 

Check  :  Substituting  4  for  a-  in  (1), 

3  .  42  -  7  •  4  -  20  =  0. 

48  -  28  -  20  =  0,  or  0  =  0. 
Substituting  —  |  for  x  in  (1), 

3(-|)^-7(-f)-20  =  0. 
.2^5  +  8^5  _  20  =  0. 

-6/  -  20  =  0,  or  0  =  0. 

A  method  of  solving  a  quadratic  equation  of  the  general 
type  in  x  by  completing  the  square  is  stated  in  the 

Rule.  Transpose  so  that  the  tei^ms  containing  x  are  in  the 
first  member  and  those  which  do  not  contain  x  are  in  the  second. 

Divide  both  ynewhers  of  the  equation  by  the  coefficient  of  x"^ 
(unless  the  coefficient  of  x^  is  -{-  i). 

Then  add  to  both  members  the  square  of  one  half  the  coeffi- 
cient of  X  (In  the  equation  just  obtained),  thus  making  the  first 
member  a  perfect  trinom,ial  square. 

118 


QUADRATIC  EQUATIONS  119 

Rewrite  the  equation,  expressing  the  first  Tnemher  as  the 
square  of  a  binomial  and  the  second  Tnemher  in  its  simplest 
form. 

Extract  the  square  root  of  both  memhers  of  the  equation  and 
write  the  sign  ±  before  the  square  root  of  the  second  member, 
thus  obtaining  two  linear  equations. 

Solve  for  X  the  equation  in  which  the  second  m^ember  is  taken 
with  the  sign  +,  dnd  then  solve  the  equation  in  which  the  second 
member  is  taken  ivith  the  sig7i  —.  The  two  results  are  the  roots 
of  the  quadratic. 

Check.  Substitute  each  result  separately  in  place  of  x  in  the 
original  equation.  If  the  resulting  equations  are  not  obvious 
identities,  simplify  until  each  becomes  one. 

Quadratic  equations  often  arise  in  which  the  first  power  of 
the  unknown  is  missing.    They  are  of  the  type  oat^  =  c.    Here 

^  =  ±  \~'    It  is  evident  that  the  solution  of -such  equations 

does  not  require  the  completion  of  the  square.    The  student 
has  solved  many  equations  of  this  type  in  Chapter  VI. 

EXERCISES 

Solve  by  completing  the  square  and  check. 

(Find  the  values  of  the  unknown  in  Exercises  10-12  correct  to 
four  decimals.) 

1.  ic2_4^_32  =  0. 

2.  2x2  + 5a; +  3  =  0. 

3.  3s2  +  8s  +  4  =  0. 

4.  3  +  5x2  =  8x.  ' 

5.  ^f  =  t-^2. 
^  x2  +  3      5.'r2-24 

7.  12 +  7x- 10x^  =  0. 

8.  39  2/  -  14  2/'  -  10  =  0.  14.  (2x  -  hf-{x  -  6)^  =  80. 


9. 

3^2 
5 

-  20  =  x^  -  30. 

10. 

x^- 

5x  +  2  =  0. 

11. 

37^2 

4 

-....\-\ 

12. 

Ix^ 

-  12  X  -  3  =  0. 

13. 

Zx"- 

2 

-3xV2  =  9. 

120        \  SECOND  COURSE  IN  ALGEBRA 


15. 

X  ^  x^  2 

3 
~4' 

16. 

1 

w  +  1      w 

1 
-3 

IT 

x  +  2     , 

3 

m 2m     .  9  _  ^ 


19. 


8-3 

X 

2x 

.r  +  5 

X 

10  s"- 

-1 

2-x 

s^ 

3-x 
-9, 

^    17.  »         ,  + =  =  2/^.      20.     „  2      Q  ,   o 

2a:  —  1      aj  —  5         ^*'  6s  —  s^  —  9        s  +  3 

After  a  student  has  mastered  the  solution  of  quadratics  by  factor- 
ing and  completing  the  square,  he  should  learn  the  formula  method 
(§  71)  and  should  use  thereafter  the  one  of  the  three  methods  which 
is  best  "adapted  to  the  problem  in  hand. 

01     _i ^  +  3        _g  r-f  2V5       2_    1 

'  x-4.      x^-6x-^^        '  '~Vl5~"^~V5' 

^^    2  (x'  +  8)      x'-l      19 

22.      ^  +  2     ^^3T  =  T'  24.  a:«  +  7a:t-8  =  0.  . 

The  equation  a:*  +  7  a:2  —  8  =  0  is  not  a  quadratic  equation,  but  it 
is  of  the  general  type  ax'^ "  +  bx^  -\-  c  =  0.  Here  x  occurs  in  but  two 
terms  and  its  exponent  in  one  term  is  twice  the  exponent  in  the 
other.  All  equations  of  this  form  can  be  solved  by  completing  the 
square. 

SDlution  :  x^  +  7  xi  ■}-  -\^-  =  8  -{-  ^^- =  ^. 

xt  +  J  =  ±  |. 

a;^  =  1  or  -  8. 
Whence  x  =  1  or  4. 

Check :  Substituting  1  for  a;  in  x^  +  7  a:^  —  8  =  0, 

1  +  7  -  8  =  0,  or  0  =  0. 
Substituting  4  for  x,  64  +  56  -  8  =  0. 

But  112  ^  0. 

Hence  the  equation  has  only  one  root,  1. 

25.  a.«-10a.t_ll  =  0.  30.  4a:^  +  i-?  =  0. 

26.  X*- 26x^  +  25  =  0.  ""  ^ 

27.  a:«-7ic8-8  =  0.  31.  2a?  -  3a;    =  2. 

^    2S.  x  +  x^-6=:0.  32.  3a:*  -  llx^  4-  6  =  0. 

29.  4a:^-7a:»  =  15.  33.  9  *' -  22  a;^ -f- 8  =  0. 


QUADRATIC  EQUATIONS  121 

34.  3it;^ +  5a;^  +  2  =  0.  38.  x'^ +  16x-^ -17  =  0. 

35.-2x^-9\'x  +  4.  =  0.  39.  ij-^ -lOy-^  +  9  =  0. 

36.  3rK-ll#-20  =  0.  40.  x"  ^  -  13  a^~  ^  =  -  36. 

37.  6x'-13x^-\-6  =  0.  41.  x^""  +  4:  -  5 cc'"  =  0. 

42.  (x^  -2xf-l(x'-2x)  =  -  12. 
Solution :  Let  x^—2x-=y. 


Substituting  y  for  x^  - 

-  2  X,  we  obtain 

?/2_7y=_12. 

Solving, 

2/  =  3  or  4. 

Then 

x2-2a;  =  3. 

Whence 

a:  =  3  or  -  i 

Also 

^-2  _  2  X  =  4. 

Whence 

X  =  1  ±  Vs. 

In  Exercises  43-48  do  not  expand  or  transpose  and  square. 
Solve  as  in  Exercise  42. 

43.  d(x'  +  3xf-l(x^  +  3x)-20  =  0. 
44.(.-i)V4(.-i)^8i.  •  :^/ 

45.  (4 7/ +  5) +2(4./ +  5)*  =  15. 

46.  ic^  +  5a;  +  3  Vx^  +  Sic  -  54  =  0.  <■ 

47.  ic2  -  2ic  -  5  Vic2_2^_4  +  2  =  0. 

48.  2  7/  (2  2/  +  1)  +  3  V8  2/'  +  12  2/  +  5  =  25  -  4  2/. 

49.  2x^  +  ax  —  Q>a^  =  0.  53.  ax^  +  hx  +  c  =  0. 

50.  9a52-2c«  =  7cl  54.  12  A-^  -  4/^^  -  5  ic^  ^  0. 

51.  2x^~llhx'  +  ^b^x  =  ().         _    2m   ,   8m'^ 

55.  -5-  +  -^^ x  =  0. 

56.  c2(cc  -  df  -  (1?{g  -  a')2  =  0. 

57.  mcc^  —  X  (m^  -\-  V)  =  —  m. 

58.  (ic  -  rf  +  (,S  -  xf  =  ?'2  +  >S2_ 

59.  -^^-^^^^:^  =  2.  60.^^--^  =  ?- 

m  —  X  m  c  x—  c      2. 


122        SECOND  COURSE  IN  ALGEBRA 

-,    a-'*  —  3  7/iic   ,    _  nx  ^^    x^  -\-2ax        x        o 

61.  +  2m  = 62.  J      -TTi  =7 

m  —  n  n  —  7n  a  —  b  2h      h 

2s  +  x      5s-- ic      s{x-{-s) 

DO.    ; — r r-  =  U. 

x  +  S  X  —  s  s^  —  x^ 

64.  2x^-{-^x  =  cx'  +  Zcx  +  ^. 
Hint.    (2  -  c)a;2  +  (5  -  3  c)a;  =  3. 
2  ,  5-3c  3 


c 


2  .  5-3c^  ,  /5-3  cV_     3       ,  25  -  30  c  +  9  c2 


x^  +  -7^ a;  + —    = + 


/5  -  3  cy^ 

(x  +  ^Z1?_£V=  49  -  42  c  +  9  c^  _  /7--3_£Y 
V      4-2  6'/      16  -  16  c  +  4  6-2  ~  \^4  -  2  c) 


2  -  c      16  -  16  c  +  4  c2 

2 


,5-3c  ,7- 3  c  , 

4  —  2  c  4  —  2  c 

65.  ic"  +  2s  -f  s^  =  2sx  +  2x.  a  +  a;      ^>  +  o^  _  5 

66.'  a;2  -  2  a;  +  1  =  ax  —  aa;^.  '  h  +  x       a  +  x~2 

67.  x^  ■^2x  +  1  =  hx^  +  Aa;.  ^,     aa?  +  b       nix  —  n 

71. = 

68.  ca;2  4-  3  X  =  2  a;2  4-  2  ex  —  2.  ^^  +  «      /ix  -  m 

69.  -^  +  -A---^  =  0.         72.  ;f  +  ^"  +  ^^l 
ic  +  ct      X  -\-  0      x  +  c  bx^  —  mic  +  n      n 

71.  Solution  by  formula.  The  standard  form  of  the  genera] 
quadratic  is  ,       ,  « 

The  student  solved  this  equation  (see  Exercise  53,  page  121) 
and  found  . 

X  = (F) 

The  value  (F)  is  a  general  result  and  may  be  used  as  a  for- 
mula to  solve  any  quadratic  equation.  The  solution  of  a  quad- 
ratic by  formula  requires  less  labor  than  by  any  other  method, 
except  for  such  equations  as  can  be  solved  by  factoring  at 
sight.  Those  who  have  considerable  exj)erience  in  algebra  sel- 
dom solve  a  quadratic  by  any  other  method  than  by  formula. 


i 


QUADRATIC  EQUATI0:NS  123 

EXAMPLES 

Solve  by  formula  and  check  : 

1.  3x''-5x  =  S. 

Solution  :  Writing  in  standard  form, 

Then  3  corresponds  to  «,  —  5  to  b,  and  —  8  to  c  in  the  general 
quadratic  ax^  ■}■  hx  +  c  =  0.    Substituting  these  values  in  (F),  where 


X  = 


.    .  -(-5)±  V25-4.3(-8) 

gives  X  =  — ^^ — ^^ 

2  '  o 


5  ±  V25  +  96      5  ±  11      8 

= = =  -  or  —  1. 

6  6  3 

Check  as  usual. 

2.   2  /^V  =  kx  +  1. 

Solution  :  Writing  in  standard  form, 

2  k^x^  -kx-l  =  0. 

Then  a  =  2k^,b=-  k,  and  c  =  -  1. 
Substituting  these  values  in  the  formula  (F), 

^_-(-k)±^(-kY-4.-2B(=T) 
2  ■2k^ 

_k±  V^^^  +8k^_k±dk_i     __i_ 

4P  -     4p     - /.^^      2  k' 

Check  as  usual. 

EXERCISES 

Solve  for  x  by  formula  and  check : 

1.  2a32-7x  +  3=:0.  6.  6x''-7rx  +  2r  =  0. 

2.  3x''-x-2  =  0.  7.  Sx^-6ax-i-2a^  =  0. 

3.  llhx  +  20Ji'  =  3x^  8.  Svi'  +  A.mx  — 7x^=0. 

4.  5x^-}-2cx  =  16c\  9.  4a:z;  -  10«V  +  3  =  0. 

10.  12y^+7v'ic-10a;2^0. 


124  SECOND  COURSE  IX  ALGEBRA 

11.  x^  -{-  Sx  =  Tnx  H-  3 7?i. 

Hint,  x^  +  (3  -  ?«)  a  —  3  m  =  0.  Then  a  =  1,  6  =  3  -  ?/i,  and  c  =  3  ?n. 
Substituting  these  values  in  (F), 

^  _  -  (3  -  m)  ±  V(3  -  m)^  -  4  ■  1(-  3"^    ^^^ 
Ji 

12.  .r^  H-  r?^  =  ca;  +  c/i.  15.  a^a;^  —  2  aa;  =  Z>V  —  1. 

13.  3a;2  _  6ca;  +  2c  =  2C.  16.  m^x^  +  mx^  +  2a;  =  4. 

14.  ?wa7^  +  hmx^  =  A;c  +  ex.        17.  tiV  +  2  tw?  =  5  /i^a;  +  10. 

18.  hkx''  -  hk  =  h^x  +  /tV. 

19.  2x'^-\-5x  =  hx^-\-Shx  +  S. 

20.  AV4-/i^-}-4a;  =  6  +  2Aa;2. 

21.  cx^  +  cma^  +  5  =  ex  +  5  (ic  +  m). 

22.  n^x  —  Snx  +  2x  =  nx^-\-6n  +  Sn\ 

23.  mV  +  4  //lic  +  7mic  +  3  /icc  =  9  x"-^  +  12  a;  —  4  //. 

PROBLEMS 

1.  Separate  20  into  two  parts,  sucli  that  the  first  shall  be 
the  square  of  the  second. 

2.  One  leg  of  a  right  triangle  is  8  feet  and  the  hypotenuse 
is  2  feet  longer  than  the  other  leg.  Find  the  other  leg,  the 
hypotenuse,  and  the  area. 

3.  The  hypotenuse   of  a  right  triangle   is  18  feet  longer 

than  one  leg  and   16   feet  longer  than  the  other.    Find  the 

three  sides. 
/ — 
/4.  The  number  of  houi-s    required  to  make  a  trip  of  112 

miles  was  6  more  than  the  rate  in  miles  per  hour.    Find  the 

rate  and  the  time. 

S".  The  Slim  of  the  reciprocals  of  two  consecutive  nmnbers 
is  |.    Find  the  numbers. 

6.  The  altitude  of  a  triangle  is  4  feet  less  than  the  base. 
The  area  of  the  triangle  is  48  square  feet.  Find  the  base  and 
the  altitude. 


QUADRATIC  EQUATIONS  125 

7.  One  leg  of  a  right  triangle  is  7  feet  shorter  than  the 
other  and  the  area  is  30  square  feet.  Find  the  three  sides  of 
the  triangle. 

8.  The  area  of  a  triangle  is  40  square  yards  and  the  base 
is  2  feet  more  than  seven  times  the  altitude.  Find  the  base  and 
the  altitude. 

9.  The  area  of  a  trapezoid  is  60  square  feet.  One  base  is 
2  feet  more  than  the  altitude  and  the  other  base  is  twicp  the 
altitude.    Find  the  bases  and  the  altitude. 

10.  One  base  of  a  trapezoid  exceeds  the  other  by  16  feet,  the 
altitude  is  2  feet  more  than  one  third  of  the  shorter  base,  and 
the  area  is  il6i  square  yards.    Find  the  bases  and  the  altitude. 

'  11^  A  requires  4  more  days  than  B  to  do  a  piece  of  work. 
If  in  working  together  they  require  8f  days,  find  the  number 
of  days  each  requires  alone. 

|^;^_  One  diagonal  of  a  rhombus  exceeds  the  other  by  4  inches. 
Find  each  if  the  area  of  th^  rhombus  is  198  square  inches. 

13.  The  radius  of  a  circle  is  21  inches.  How  much  must  it 
be  shortened  so  as  to  decrease  the  area  of  the  circle  770  square 
inches  ?    (Use  tt  =  ^-^-.) 

|;14.|ln  selling  an  article  at  18  dollars  a  merchant  gained  a 
per  cent  5  greater  than  the  number  of  dollars  the  article  cost. 
Find  the  cost  in  dollars  and  the  gain  per  cent. 

—  15.  From  a  cask  full  of  pure  wine  5  gallons  are  drawn  off. 
The  cask  is  thei>  filled  by  adding  pure  water,  and  again  5  gal- 
lons are  drawn^ofi. -.11  36  i^er  c^nt  of  the  mixture  is  water, 
how  many  gallons  does  the  cask  cbntain  ? 

16.  A  printed  page  has  15  more  lines  than  the  average  num- 
ber of  letters  per  line.  If  the  number  of  lines  is  increased  by 
15,  the  number  of  letters  per  line  must  be  decreased  by  10  in 
order  that  the  amount  of  matter  on  the  two  pages  may  be  the 
same.    How  many  letters  are  there  on  the  jjage  ? 


12G  SECOND  COURSE  IN  ALGEBRA 


\1 


i 


\lTJ  A  sum  of  |5000  is  put  at  interest.  At  the  end  of  each 
year  the  yearly  interest  and  |300  are  added  to  the  investment. 
If  at  the  beginning  of  the  third  year  the  investment  amounts 
to  $6236,  find  the  rate  of  interest  that  the  investment  bears. 

.'  18. '  The  cost  of  an  outing  was  $36.  If  there  had  been  2 
more  in  the  party,  each  would  have  been  required  to  pay  $3 
less.    Find  the  number  in  the  party. 

19.  Two  bodies,  A  and  B,  move  on  the  sides  of  a  right  trian- 
gle. A  is  now  123  feet  from  the  vertex  and  is  moving  away 
from  it  at  the  rate  of  239  feet  per  second.  B  is  239  feet  from 
the  vertex  and  moves  toward  it  at  the  rate  of  123  feet  per  sec- 
ond.   At  what  time  (past  or  future)  are  they  850  feet  apart  ? 

20.  The  dimensions  of  a  rectangular  box  in  inches  are  ex- 
pressed by, three  consecutive  numbers.  The  surface  of  the  box 
is  292  square  inches.   Find  the  dimensions. 

[21./ A  three-inch  square  is  cut  from  each  corner  of  a  square 
piece  of  tin.  The  sides  are  then  turned  up  and  an  open  box  is 
formed,  the  volume  of  which  is  300  cubic  inches.  Find  in 
inches  the  side  of  the  piece  of  tin. 

22.  A  piece  of  tin  is  10  inches  by  12  inches.  From  each 
corner  a  square  is  cut  whose  side  is  x  inches.  The  sides  are 
turned  up  and  an  open  box  is  formed.  Show  that  its  volume 
is  4a;3-44ic2_^120a;. 

23.  Now  a  certain  value  of  x  gives  for  the  box  in  Exercise 
22  the  greatest  possible  volume.  That  value  is  one  root  of  the 
equation  12  x^'-SSx  +  120  =  0.    Find  the  value  of  x. 

24.  A  rectangular  box  is  8  inches  long.  Its  volume  is  192 
cubic  inches  and  the  area  of  its  six  faces  is  208  square  inches. 
Find  the  three  dimensions. 

25.  A  messenger  leaves  the  rear  of  an  army  28  miles  long  as 
it  begins  its  day's  march.  He  goes  to  the  front  and  at  once 
returns,  reaching  the  rear  as  the  army  camps  for  the  night. 
How  far  did  he  travel  if  the  army  went  28  miles  during  the 
day? 


QUADRATIC  EQUATIONS 


127 


26.  If  AB  in  the  adjacent  figure  is  a  tangent  to  the  circle 
and  BD  is  any  secant,  Ab''  =  BC  •  BD.  Find  BC  if  AB  =  12 
and  CD  =  20. 

"^27.  How  high  is  a  mountain 
which  can  just  be  seen  from  a  point 
on  the  surface  of  the  sea  80  miles 
distant  ?  (Use  3960  miles  for  the 
radius  of  the  earth.) 

28.  Find  the  distance  a  man  can 
see  in  a  straight  line  over  a  smooth 
lake,  if  his  eye  is  6  feet  above  the 
level  of  the  water. 

29.  Two  lighthouses  on  opposite  shores  of  a  bay  are  150  and 
250  feet  respectively  above  the  water.  If  the  light  from  one 
can  just  be  seen  from  the  other,  find  the  distance  in  miles 
between  them. 

30.  A  ship  is  31  miles  from  a  lighthouse  which  is  250  feet 
above  the  water.  How  high  above  the  water  is  the  ship's  flag 
if  it  can  just  be  seen  from  the  lighthouse  ? 

^  31.  A  stone  dropped  from  a  balloon  which  was  passing  over 
a  river  struck  the  water  12  seconds  later.  How  high  was  the 
balloon  at  the  time  the  stone  was  dropped  ? 

Hint.  The  distance  S  through  which  a  body  falls  from  rest  in  t 
seconds  is  given  by  the  equation  S  =  ^  (^  =  32  feet,  approximately). 

32.  A  man  drops  a  stone  over  a  cliff  and  hears  it  strike  the 
ground  below  13  seconds  later.  If  sound  travels  1120  feet  per 
second,  find  the  height  of  the  cliff. 


CHAPTER  IX 

IRRATIONAL  EQUATIONS 

72.  Definitions  and  typical  solutions.  An  irrational  or  radical 
equation  in  one  unknown  is  an  equation  in  which  the  unknown 
letter  occurs  in  a  radicand. 

Thus  3  a:  +  2  -v^  =  16,  Vl  -  x  +  Vx  +  3  =  2,  and  "v^x^  -  8  =  0 
are  irrational  equations.  Also  any  equation  in  which  the  unknown 
occurs  with  a  fractional  exponent  is  irrational. 

The  following  examples  illustrate  the  method  of  solution  for 
some  of  the  more  simple  irrational  equations. 

EXAMPLES 


1.  Solve  V2x-5  -3  =  0. 

Solution  :  Transposing,  V2  x  —  5  =  3. 
Squaring  both  members,  2  a:  —  5  =  9. 
Solving,  X  =  7. 

Check :  Substituting  7  for  x  in  the  original  equation, 

VH-5  -3  =  0. 
Whence  3-3  =  0. 

In  irrational  equations  it  is  understood  that  each  radical 
expression,  not  preceded  by  the  sign  ±,  is  to  have  one  sign  and 
only  one  ;  therefore  each  radical  will  have  one  value  and  only 
one.  That  value  is  the  principal  root  of  the  radical.  This  fact 
is  of  importance  in  checking. 

2ic-l.  (1) 

ix  +  1.  (2) 


2. 

8 

Solve  2  -Js  x^ 
►iution :  Transpo 

.r-2. 

-2: 

Sc 

>sing,  2  y^8  x^  + 

19a:2 
2 

128 


J 


IRRATIONAL  EQUATIONS  129 

12^  +  1.    (3) 


Cubing  each  member  of  (2), 

64  x^  +  76  a:2  z 
Transposing  and  collecting, 

28  a;2  -  12  a;  -  1  = 
Factoring,           (2  x  -  1)  (14  x  +  1)  = 
Therefore                                           x  - 

=  64:x^  +  48 
=  0. 

Check  :  Substituting  1  for  x  in  (1), 

2  -^1  4-  -V-  -  1  -  2  = 
2-1-3  = 
3-3z 
Substituting  —  J^  for  x  in  (1), 

=  1-1. 

=  0. 
=  0- 

2^8(-tV)'  +  ¥(-tV)'  +  ^-2  = 

=  -4-1. 

o-      1  •  o  /  ,    5  \      13  8         -  8       -  8 

Simplying,  2(^+  -j --=--,  or  -^  =  — . 

It  is  easily  possible  to  write  a  statement  involving  radical  expres- 
sions  which  has  the  form  of  an  equation,  but  is  not  one.  Thus 
-\x  +  1  +  Va:  +  3+1  =  0  looks  like  an  equation,  but  no  value  of 
X  can  satisfy  it.  A  little  closer  inspection  shows  that  the  statement 
asserts  that  the  sum  of  three  positive  numbers  is  zero,  a  condition 
clearly  impossible.  Statements  like  the  one  given  are  often  called 
"impossible  equations,"  though,  strictly  speaking,  they  are  not  equa- 
tions at  all.  In  the  attempt  to  solve  an  apparent  equation  one  may 
resort  to  the  usual  methods  of  solution  and  obtain  a  result  which 
will  not  satisfy  the  original  statement.  Not  until  one  tries  to  verify 
the  result  is  the  falsity  of  the  original  statement  discovered. 

3.  Solve  1  +-  Vic  +  2  =  V5,  (1) 
Solution  rTransposing,        1  —  -y/x  =  —  V-f  +  2.                        -     (2) 

Squaring  (2),  1  -  2  V^  +  a;  =  x-  +  2.  ■  (3) 

Transposing  and  collecting, 

-  2  V^  =  1.  (4) 

Squaring  (4),  4  a;  =  1,  or  a;  =  ^.  (5) 

Check :  Substituting  ^  for  x  in  (1), 

i  +  VfT2=  +  V|. 

1  +  I  =  +  i,  or  I  =  1,  which  is  false. 

It  is  fairly  certain  that  the  student  did  not  see  that  the  state- 
ment (1)  is  false  until  the  attempt  was  made  to  verify  the  result. 
It  appears,  then,  that  the  method  of  solution  may  give  a  result 
which  is  not  a  root. 


130 


SECOND  COURSE  IN  ALGEBRA 


4.  Solve  Va;-l  +  V3a;4-l-2  =  0. 

(1) 

Solution  :  Transposing,    V3  a;  +  1  =  2  —  -y/x  —  1. 

(2) 

Squaring  both  members  of  (2), 

3x  +  l  =  4-4V:c-l  +  x-l. 

(3) 

Transposing  and  collecting, 

2  a;  -  2  =  -  4  Vx  -  1. 

(4) 

Dividing  (4)  by  2,                    a;  _  i  ^  _  o  Vo:  -  1. 

(5) 

Squaring  both  members  of  (5), 

a;2-2x  +  i  =  4a:-4. 

(6) 

Transposing,                  x^  -  6  a;  +  5  =  0. 

(7) 

Factoring,                  {x  -  1)  (a:  -  5)  =  0. 

Therefore                                          a:  =  1  or  5. 

Check  :  Substituting  1  for  x  in  (1), 

Vl  -  1  +  V3  +  1  -  2  =  0. 

0  +  2-2  =  0. 

Therefore'  1  is  a  root  of  (1). 

Substituting  5  for  x  in  (1), 

Vo  -  1  +  Vl5  +  1-2  =  0. 

2  +  4-2  =  0, 

or                                                            4  =  0  ;  but  4  ;zt  0. 

Therefore  5  is  not  a  root  of  (1).  It  was  introduced  by  the  process 
of  squaring  each  member  of  equation  (5).  This  process  does  not 
necessarily  introduce  a  root.  Thus  1  is  a  root  of  each  of  the  equations 
(1)  to  (7),  and  while  5  is  a  root  of  (6)  and 
(7),  it  is  not  a  root  of  (5),  as  may  be  veri- 
fied by  substitution.  Further,  (5)  was  ob- 
tained by  squaring  (2),  yet  neither  the  root  1 
nor  the  root  5  was  introduced  at  that  point. 

Just  what  did  happen  in  the  course 
of  the  preceding  solution  is  shown  in 
the  adjacent  figure,  where  equations 
(1),  (5),  and  (7)  are  solved  graphically. 

The  graph  shows  the  changes  in  the 
function  due  to  squaring.  It  appears 
that  the  root  1  is  common  to  (1),  (5), 
and  (7),  while  the  root  5  is  extraneous 
to  (1)  and  (5). 


If 

1 

L 

/  / 

jb 

ji 

(5)y  / 

71 

7-  t 

-J         l— 

r            t       4l4,Z. 

7     3riz: 

i^^l^ 

[ly^   M 

\V     t 

\r          r 

\       U            ^' 

\    i 

\      T 

SZ 

^ 

1 

IRRATIONAL  EQUATIONS 


131 


As  we  have  seen,  the  solution  of  (1)  leads  to  the  quadratic 
x^  —  Q  X  +  5  =  0.  Since  (1)  and  (5)  have  the  root  1  but  not  5,  it  is 
obvious  that  with  some  radical  equations  one  may  resort  to  squaring 
once  without  introducing  an  extraneous  root. 

Equation  (1)  is  typical  of  many  radical  equations  which,  when 
solved  by  rationalizing,  give  the  roots  not  only  of  the  original  equa- 
tion, but  also  of  such  equations  as  may  be  derived  from  it  by  giving 
each  radical  therein  the  sign  ± . 

It  will  be  seen  from  the  next  example,  also,  that  the  process  of 
rationalization  does  not  necessarily  introduce  extraneous  roots. 


5.  Solve  V^  +  2  +  V3  -  X  =  3. 

(1) 

Solution  :  Transposing,            V^  —  a;  =  8  -  V^'  +  2. 

(2) 

Squaring  (2),                                 3-x  =  9-6^x  +  2+x-{-2. 

(^) 

Transposing  and  collecting,  —  2x  —  8=— Q  -\/x  +  2. 

(4) 

(4)  ^  -  2,                                       X  +  4  =  8  Vx  +  2. 
Squaring  (5),                    x^  +  8  ar  +  16  -:  9  x  +  18. 
Transposing  and  collecting,  x^  —  x  —  2  =  0. 

(5) 
(6) 
(7) 

Factoring,  (x  -2)(x  +  l)  =  0. 

Therefore  a:  =  2  or  —  1. 

Check :  Substituting  2  for  x  in  (1), 

V2  +  2  +  V3  -  2  =  8,  or  2  +  1  =  3. 
Substituting  —  1  for  x  in  (1), 

V-l+"2  +  Va+l  =  3,  or  1  +  2  =  3. 
Therefore  equation  (1)  has  two  roots,  2  and  —  1. 


The  graphical  solution 
of  (1),  (5),  and  (7)  gives 
curves  I,  II,  and  III  re- 
spectively of  the  adjacent 
figure.  These  graphs  show 
the  change  in  the  function 
'with  each  resort  to  squar- 
ing. Curves  I,  II,  and  III 
intersect  the  x-axis  at  the 
same  points,  showing  that 
the  roots  2  and  —1  are 
common  to  (1),  (5),  and  (7), 


t 

/ 

\ 

/ 

\ 

/ 

\ 

n 

/ 

I 

\ 

1 

\ 

' 

1 

^ 

' 

-X- 

^ 



__ 

1 

^ 

/ 

V 

~o 



TTI 

^ 

^ 

s, 

J 

/ 

\ 

/ 

V 

/ 

f 

\ 

T 

TL 

/ 

I> 

u 

_j 

r 

132  SECOND  COURSE  IN  ALGEBRA 

It  should  be  clear  from  the  preceding  examples  that  we  cannot 
determine  the  number  of  roots  of  a  given  radical  equation  witliQut 
solving  it ;  nor  can  we  predict  whether  the  given  statement  involv- 
ing radicals  is  an  equation.  Results  obtained  are  roots  if  they  satis/// 
the  original  statement,  and  not  otherwise. 

The  method  of  solving  a  radical  equation  may  be  stated 
in  the 

Rule.  Transpose  the  terms  so  that  one  radical  expression 
{the  least  simjde  one)  is  the  only  term  in  one  member  of  the 
equation. 

Next  raise  both  members  of  the  resulting  equation  to  the  same 
power  as  the  index  of  this  radical. 

If  radical  expressions  still  remain,  repeat  the  two  preced- 
ing operations  until  an  equation  is  obtained  which  is  free  from 
radicals.    Then  solve  this  equation. 

Check.  Substitute  the  values  found  in  the.  original  equa- 
tion and  reduce  the  resulting  radicals  to  their  simplest  form. 
Whenever  the  radicals  are  rational  simplify  by  extracting  the 
roots  indicated.  Never  simplify  by  raising  both  members  of 
the  equation  to  any  power,  for  extraneous  roots  introduced  by 
that  process  would  not  then  be  detected. 

Finally,  reject  all  extraneous  roots. 

EXERCISES 

Solve,  check  results,  and  reject  all  extraneous  roots : 
^1.  V^T3  =  8.  >(9^)^W_+4. 

1    2.  V2ic- 6  +  4  =  7.  9.   ^5n'  +  19  +  n  =  7. 

v3.  3V2^38-7  =  17.      ^10-  (3^-4)*  +  (4a.  +  3)*  =  0. 

''4.  -v^S^ITi  =  2.  11.  4Vv^-^-4  +  3  =  15. 

^  5.  (7x  +  15)*  +  18  =  17.        12.  V^T~4  =  ■\/x'' -  5  x  +  6. 

^  6.  2  -v^3  71  -  25  4-3  =  7.         13.  -y/x-^-l  =  V^c  +  1. 
^xVs^  =  18  V27^.  14.  (s  -  2f  =  5*  +  2*. 

157  3  Vr  +  1  -  2  Vr  +  3  =  V2r  +  4  -  V>' +  3  -f  2  V7+1, 


/  IRRATIONAL  EQUATI0:N^S  133 

V 

16.  V2x  -3  +  3Vic-5  =  V3ic-8  +  2Vic-5. 

17.  V4a;  -  12  +  V5a.-  -  2  +  V9ic  -  14  =  0. 

18.  V4.r  -  12  -  V5x  -  2  -  V9x  -  14  =  0. 


19.  V4  ;z;  -  12  +  V5  a;  -  2  -  V9  £c  -  14  =  0. 

20.  V4a2  -  12  -  V5a;  -  2  +  V9ic  -  14  =  0. 


21.  ^^+^"  =  3.  24,^^^+15+^4 


22. 


4n-2  V4-X        Vx  +  16      2. 

^^A/a  V2a-  +  4«.      25.  5  r  -  13  v^  +  6  =  0. 


V2ic-a  3V7t  26.  7x-3ic2^-2ic§  =  0. 


23.  ^^^-3      ^-^^_0.      '  27.  0-  +  ^)^      (r  +  3)^_    2_ 
71^  '^  '  '  (r  +  3)^      (r  +  5)*      V3 

28.  V7  +  4.x  +  3V2a;2  +  5ic  +  7-3=:0."^  ^-.  H  '] 


^ 


29.  VlT  +  2  V3  +  s  +  Vs  +  7  -5=0. 

30.  4£c2  =  10t  +  10-2V4^-^-10;7;-2. 


31.  3  irr  =  6  V3  >m^  -  m  -  6  +  w^.  +  22. 

32.  V^'  +  15  +  -s/x  —  24  —  Va.'  —  13  =  V^. 


33.  Solve  for  I  and  r/,  ^  =  tt^  - •     34.  Solve  ioY  t,  s  —  ^- 

yff  ^ 

35.  If  «  =  —  V2  and  K  =  2  R'^,  express  K  in  terms  of  «. 

36.  If  K  =  *— -  V 3  and  a  =  -^  Vs,  express  K  in  terms  of  «. 


37.  If  A^  =  2  r-  V2  and  a  —  j  v  2  +  V2,  express  K  in  terms 
of  rr. 

38.  If  A'  =  3  r-  and  o!-  =  -^  v  2  +  V3,  express  A  in  terms  of  a. 

39.  The  perimeter  and  the  area  of  a  certain  square  exceed 
the  perimeter  and  area  of  a  second  square  by  72  feet  a«d  900 
square  feet  respectively.    Find  the  side  of  each  square. 


134  SECOND  COURSE  IX  ALGEBRA 

40.  If  a  bullet  is  fired  vertically  upward,  the  least  velocity 
V  which  it  may  have  so  that  it  will  never  return  to  the  earth 
is  given  by  the  equation  V  =  V2  gli.  (g  =  32  feet  per  second, 
R  =  4000  miles.)  Find  the  velocity  in  miles  per  second  to  the 
nearest  whole  number. 

41.  The  greatest  distance  x  (in  feet)  that  a  ball  can  be  thrown 
with  velocity  v  (in  feet  per  second)  across  a  level  field  is  given 
by  one  root  of  the  equation  .976  v^x  —  gx^  =  0.  (g  =  32.)  Under 
the  conditions  just  stated  a  ball  is  thrown  with  a  velocity  of 
100  feet  per  second.  How  far  from  the  thrower  does  it  strike 
the  ground  ? 

42.  The  greatest  distance  a  baseball  has  been  thrown  is 
426  feet  6^  inches  (Sheldon  Lejeune,  October  10, 1910).  With 
what  velocity  did  it  leave  the  thrower's  hand  ?  (This  velocity 
is  called  the  initial  velocity.) 

43.  Determine  the  initial  velocity  from  the  data :  («)  A  La- 
crosse ball  has  been  thrown  497  feet  7^  inches  (B.  Quinn, 
1902).  (b)  The  record,  distance  for  the  16-pound  shot  is  51  feet 
(Ealph  Kose,  1909).  (c)  The  16-pound  hammer  has  been  thrown 
184  feet  4  inches  (John  Flanagan,  1910).  (d)  A  football  has 
been  kicked  a  distance  of  200  feet  (W.  P.  Chadwick,  1887). 


CHAPTER 

GRAPHS  OF  QUADRATIC  EQUATIONS  IN  TWO  VARIABLES 

73.  Graph  of  a  quadratic  equation  in  two  variables.  Before  solv- 
ing graphically  a  quadratic  system,  the  method  of  graphing  one 
quadratic  equation  in  two  variables  must  be  clearly  understood. 


EXAMPLES 


1.  Construct  the  graph  oi  x^  =  Z  y. 

Solution  :  Solving  the  equation  for  x  in  terms  oi  y,  x  —^  V3^. 
We  now  assign  values  to  y  and  then  compute  the  approximate 
corresponding  values  of  x.    Tabulating  the  results  gives : 


.'/  = 

9 

4 

3 

2 

1 

0 

-1 

Any  negative  value 

,.= 

±5.19 

±3.4G 

±3 

±2.44 

±1.73 

0 

-V^ 

Imaginary 

Using  an  a:-axis  and  a  ^-axis 
as  in  graphing  linear  equations, 
plotting  the  points  correspond- 
ing to  the  real  numbers  in  the 
table,  and  drawing  the  curve 
determined  by  these  points,  we 
obtain  the  graph  of  the  adja- 
cent figure.  Since  ?/  is  a  func- 
tion of  X,  the  2/-axis  corresponds 
to  the  function  axis.  The  curve 
is  a  parabola.  A  similar  curve 
was  always  obtained  in  Chapter 
VII  for  the  graph  of  a  quadratic 
function  of  one  variable. 


The  graph  of  any  equation  of  the  form  y^  =  ajf  is  a  parabola. 

135 


Y 

, 

lP 

AR 

AB< 

DU 

\ 

/ 

\ 

GR 

AP 

H 

/ 

\ 

x= 

=  3 

/ 

\ 

\ 

/ 

\ 

/ 

V 

/ 

\ 

s. 

/ 

, 

\ 

Sj 

^ 

A 

-X- 

0 

i 

_ 

/' 

130 


SECOND  COURSE  IN  ALGEBRA 


2.  Graph  the  equation  xy  -\-  S  =  0. 

Solution :  Solving  for  y  in  terms  oi  x,  y  = 

Assigning  values  to  x  as  indicated  in  the  following  table,  we  then 
compute  the  corresponding  values  of  y. 


x  = 

-6 

-5 

-4 

-3 

-2 

-1 

-1 

^ 

1 

^ 

3 

4 

5 

6 

8 

?/  = 

1 

8 

2 

§ 

4 

8 

¥ 

-¥- 

-8 

-4 

-1 

-2 

-f 

-1 

-1 

> 

^ 

/ 

/ 

/ 

/ 

, 

HYP 

ER 

JO 

-A 

y 

I 

^ 

^ 

-> 

t 

0 

I     ■ 

^ 

■ 

'" 

". 

GRARH  < 

DF 

y 

X 

^  + 

B= 

0 

/ 

j 

1 

1 

t 

" 

L_ 

Proceeding  as  before  with  the  numbers  in  the  table,  we  obtain  the 
two-branched  curve  of  the  above  figure,  which  does  not  touch  either 
axis.    The  curve  is  called  an  hyperbola. 

The  graph  of  any  equation  of  the  form  jry  =  jK"  is  an  Jiyper- 
hola.  The  curve  for  xy  —  K  (K  =  any  constant)  is  always  in 
the  same  general  position.  That  is,  if  K  is  positive,  one 
branch  of  the  curve  lies  in  the  first  quadrant  and  the  other 
branch  in  the  third.  If  K  is  negative,  one  branch  lies  in  the 
second  quadrant  and  the  other  in  the  fourth. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


137 


3.  Graph  the  equation  x^ 


16. 


Solution  :  Solving  for  y  in  terms  of  x,  y  =  ±  -vK)  —  x^. 

Assigning  values  to  x  as  indicated  in  the  following  table,  we  com- 
pute the  corresponding  approximate  values  of  y. 


x  = 

5 

-4 

-3 

-2 

-1 

0 

1 

2 

3 

4 

5 

?/  = 

±3V~1 

0 

+  2.G4 

±3.46 

+  3.87 

±4 

±3.87 

±3.46 

±2.G4 

0 

±3V^ 

For  values  of  x  numerically  greater  than  4,  y  is  imaginary.    The 
points  corresponding  to  the  pairs  of  real  numbers  in  the  table  lie  on 
the  circle  in  the  adjacent  figure. 
The  center  of  the  circle  is  at 
the  origin  and  the  radius  is  4. 

Further,  the  graph  of  any 
equation  of  the  form 

A:2  +  z/2  =  r2 

is  a  circle  whose  radius  is  r. 
This  can  be  proved  from  the 
right  triangle  PKO.  If  P  rep- 
resents any  point  on  the  circle, 
OK  equals  the  x-distance  of 
P,  KP  equals  the  ?/-distance, 
and  OP  equals  the  radius.    Now  OK^  +  KP^  =  0P^\  that  is, 


, 

^z 

-^ 

S 

/ 

t\ 

s 

/ 

1 

,    « 

:iR 

CLE 

, 

^   1! 

\ 

— 

K 

0 

. 

I 

V' 

RA 

r.^ 

OF 

\ 

s. 

>- 

/ 

S 

^ 

y 

f 

X^+lf 


It  follows,  then,  that  the  graphs  oi  x^  -\-  y^  =  9 


and  x^  -\-  y^  =  S  are  circles  whose  centers  are  at  the  origin 
and  whose  radii  are  3  and  Vs  respectively.  Hereafter, .  when 
it  is  required  to  graph  an  equation  of  the  form  x^  -\-  y^  =  r^, 
the  student  may  use  compasses,  and,  with  the  origin  as  the 
center  and  the  proper  radius  (the  square  root  of  the  constant 
term),  describe  the  circle  at  once. 

In  all  of  the  graphical  work  which  follows  it  is  expected 
that  the  student  will  save  time  by  obtaining  from  the  curve  on 
page  85,  or  from  the  table  on  page  262,  the  square  roots  or 
cube  roots  which  he  may  need. 


138 


SECOND  COURSE  IN  ALGEBRA 


4.  Graph  the  equation  16x^  +  9y^  =  144. 


Solution  :  Solving  for  y  in  terms  of  a:,  y  =  ±  |  s/d  —  x^. 

Assigning  values  to  x  as  indicated  in  the  following  table,  we  com- 
pute the  corresponding  approximate  values  of  y. 


x= 

-4 

-3 

-2 

-1 

0 

+  1 

+  2 

+  3 

+  4 

y= 

.fV^T 

0 

±2.98 

±3.77 

±4 

±3.77 

±2.98 

0 

±.fV^ 

— ■■  '    y' 

^^--^ 

^        ^      s 

Z                  S, 

^                        ^ 

'1                      t 

/gf^a^o^                    \ 

/l    x%  9y=l  44  ' 

i:± :.x 

L                   : 

\                          ELLIPSE     / 

i                                             t 

^  ^                  _r 

\                 z 

s     ^dz 

-^ 

For  values  of  x  numerically 
greater  than  3,  ^/  is  imaginary.  The 
points  corresponding  to  the  real 
numbers  in  the  table  lie  on  the 
graph  of  the  adjacent  figure.  The 
curve  is  called  an  ellipse. 

The  graph  of  any  equation 
of  the  form  of  ax^  -\-  by^  =  c, 
in  which  a  and  b  are  unequal 
and  of  the  same  sign  as  c,  is 
an  ellipse. 

Note.  These  three  curves,  the 
ellipse,  the  hyperbola,  and  the  pa- 
rabola, were  first  studied  by  the 
Greeks,  who  proved  that  they  are  the  sections  which  one  obtains 
by  cutting  a  cone  by  a  plane.  Not  for  hundreds  of  years  after- 
wards did  any  one  imagine  that  these  curves  actually  appear  in 
nature,  for  the  Greeks  regarded  them  merely  as  geometrical  figures, 
and  not  at  all  as  curves  that  have  anything  to  do  with  our  every- 
day life.  One  of  the  most  important  discoveries  of  astronomy  was 
made  by  Kepler  (1571-1630),  who  showed  that  the  earth  revolves 
around  the  sun  in  an  ellipse,  and  stated  the  laws  which  govern  the 
motion.  Those  comets  that  return  to  our  field  of  vision,  periodically 
also  have  elliptic  orbits,  while  those  that  appear  once,  never  to  be 
seen  again,  describe  parabolic  or  hyperbolic  paths. 

The  path  of  a  ball  thrown  through  the  air  in  any  direction,  except 
vertically  upward  or  downward,  is  a  parabola.  The  approximate  pa- 
rabola which  a  projectile  actually  describes  depends  on  the  elevation 
of  the  gun  (the  angle  with  the  horizontal),  the  quality  of  the  powder, 
the  amount  of  the  charge,  the  direction  of  the  wind,  and  various  other 
conditions.    This  makes  gunnery  a  complex  problem. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


139 


EXERCISES 

Construct  the  graphs  of  the  following  equations  and  state 
the  name  of  each  curve  obtained : 

1.  ic2  =  2y. 

2.  if^lx 

3.  x"-  ^y^-. 


x^  +  if 
x^  —  if' 


=  0. 
36. 
12. 


=  2^.1^' 


6.  xy  =  8.    WV'  J^  ^ 

7.  .,  =  -12.^'     ^^f% 

8.  9cc2_^16  2/'  =  144.  ^^^ 

9.  16a;2-9^2  =  144.  V/^ 
10.  25ic2  +  9  2/'=:225. 


74.  Graphical  solution  of  a  quadratic  system  in  two  variables. 

That  we  may  solve  a  system  of  two  quadratic  equations  by  a 
method  similar  to  that  employed  in  §  38  for  linear  equations 
appears  from  the  following 


1.  Solve  graphically 


EXAMPLES 

2x  +  y  =  l,  •     .         (1) 

/  +  4^  =  17.  (2) 

Solution :  Constructing  the  graphs  of  (1)  and  (2),  we  obtain  the 

straight  line  and  the  parabola  shown  in  the  adjacent  figure.    There 

are  two  sets  of  roots  corresponding 

to  the  two  points  of  intersection, 

which  are : 


-2, 

5, 


B 


2, 


3. 


Note.  If  the  straight  line  in  the 
adjacent  figure  were  moved  to  the 
right  in  such  a  way  that  it  always 
remained  parallel  to  its  present 
position,  the  points  A  and  B  would 
approach  each  other  and  finally 
coincide.  The  line  would  then  be 
tangent  to  the  parabola  at  the  point 
X  =  ^,  y  =  1. 

Were  the  straight  line  moved  still  farther,  it  would  neither  touch 
nor  intersect  the  parabola  and  there  would  be  no  graphical  solution. 

An  illustration  of  these  two  conditions  is  given  by  the  graphical 
solution  of  Exercises  8  and  9,  page  142. 


1 

_^    V 

^5 

A^^ 

^    N^ 

^2          ^^ 

\       3 

yi                                   V                  L-X 

^                                         0\       2     3       /      ^^ 

^  z 

V 

^^\(i^ 

^-^ 

Y' 

140 


SECOND  COURSE  IN  ALGEBRA 


2.  Solve  graphically 


(1) 

(2) 


Solution :  Constructing  the  graphs  of  (1)  and  (2),  we  obtain  the 
hyperbola  and  the  parabola  of  the  following  figure.    There  are  four 


c 

\ 

p> 

/ 

\ 

A 

^ 



V^ 

' 

\ 

R^ 

/ 

y^ 

\ 

/ 

, 

/' 

^ 

\ 

0 

2 

^ 

/ 

\ 

/ 

f^ 

> 

\j 

/ 

K 

— 

iSL 

/ 

^ 

) 

■ 

" 

" 

rx  =  3.7,  rx=-2.7,  rx=-2.7, 

^12,  =  3.1.        ^12^  =  1.8.  ^\y=-1.8. 

3.  Solve  graphically  j    2      2r  —    j 


sets  of  roots  corresponding  to  the  four  points  of  intersection,  which 
are  approximately 

r:.  =  3.7, 

ny=-3.i. 

(1) 
f  =  ie.  (2) 

Solution :  The  graphs  (1)  and  (2)  are  the  circle  and  hyperbola  of 
the  figure  on  page  14r.  These  curves  have  no  real  points  of  inter- 
section. There  are,  however,  four  sets  of  imaginary  roots.  Sub- 
tracting equation  (2)  from  (1)  gives  1]f-  =  —  1,  whence  y  =  ±  V— ^, 
an  imaginary  expression.  Adding  (1)  and  (2)  gives  2  a;^  =  25,  whence 
X  —  ±  ^  V2,  a  real  expression.  Using  the  double  sign  before  each 
radical  gives  the  four  sets  of  imaginary  roots  : 

ra;  =  |V2,  +IV2,  -|-\^,  -|V2. 

U  =  V^,     -V=|,     -v^,     +v^. 

It  can  be  shown  that  these  sets  of  imaginary  roots  correspond  to 
tiie  intersections  of  what  may  be  termed  the  imaginary  branches  of  the 


GRAPHS  OF  QUADRATIC  EQUATIONS 


141 


curves.  These  branches  may  be  represented  as  lines  not  in  the  same 
plane  as  the  real  branches  but  in  a  plane  passing  through  the  x-axis 
perpendicular  to  the  plane  determined  by  the  a;-axis  and  the  y-axis. 


\                 ^                 ^ 

V(2)                                                                   (.2)/ 

^v                               / 

^               (T         Z 

L  -.'^^r^v-  -i 

\T    T  X  t 

Ji    J  t 

i  {     °'  2 

7  S         ^  V 

/    ^^-^     \ 

7                          V 

y                          V 

y                           ^. 

y                             ^ 

/ 

Though  the  subject  is  not  difficult,  even  a  simple  presentation  of  this 
method  of  constructing  imaginary  graphs  is  wholly  beyond  the  scope 
of  this  book.  The  essential  point  to  be  grasped  now  is  that  real 
roots  correspond  to  real  intersections,  and  imaginary  roots  correspond 
to  no  intersections  of  real  graphs. 

Note.  In  equation  (1),  page  140,  a  greater  number  in  place  of  9 
would  give  a  larger  circle  than  the  one  in  the  figure,  and  it  would  be 
easy  to  find  a  number  to  replace  9  such  that  the  resulting  circle 
would  just  touch  the  hyperbola.  AVere  a  still  greater  number  used, 
the  circle  obtained  would  intersect  the  other  curve.  These  varying 
conditions  would  result,  respectively,  in  (a)  no  set  of  real  roots, 
(b)  two  sets  of  real  roots,  (c)  four  sets  of  real  roots. 


Examples  1,  2,  and  3  partially  illustrate  the  truth  of  the 
following  statement : 

If  in  a  system  of  two  equations  in  two  variables  one  equa- 
tion is  of  the  mth  degree  and  one  of  the  nth,  there  are  usually 
mn  sets  of  roots  (real  or  imaginary)  and  never  more  than  mn 
such  sets. 


142  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

If  possible,  solve  graphically  each  of  the  following  systems  ; 

x'^  =  4:y,  x^  +  f  =  4:,  y^=^x, 

'  x-\-Sy  =  5.  '  x  +  y  =  S.  x^-{-9i/  =  9. 

x'-\-f  =  25,  x'  +  f  =  25,  2/2  +  4.^  =  17, 

'  a;  -  2?/  =  10.  x^-f=16.  2ic  +  2/  =  9. 

a.2  +  /  =  16,  x'^f  =  16,  f-{-^x  =  17, 

'''  x^-\-f  =  9.  x^-f=^25.  2x  +  7/  =  12. 

x'  +  f  =  16,  2^_^V8  =  0, 

a^  +  ,/-2x  =  8.  /  =  :r8  -  9i». 


CHAPTER  XI 

SYSTEMS  SOLVABLE  BY  QUADRATICS 

75.  Introduction.  The  general  equation  of  the  second  degree 
in  two  variables  is  ax^  +  hif'  +  cxy  ■\-dx-\-  ey  +/=  0.  To  solve 
a  pair  of  such  equations  requires  the  solution  of  an  equation 
of  the  fourth  degree.  Even  the  solution  of  a?^  -|-  ^  =  5  and 
?/2  -|-  X  =  3  requires  the  solution  of  a  biquadratic  equation. 
In  fact,  only  a  limited  number  of  systems  of  the  second  degree 
in  two  variables  are  solvable  by  quadratics.  The  student 
should  note  that  he  can  solve  graphically  for  real  roots  any 
system  of  quadratic  equations,  provided  the  terms  have  nu- 
merical coefficients.  The  algebraic  solution  of  such  systems 
will  be  possible  for  him  only  after  further  study  of  algebra. 

76.  Linear  and  quadratic.  Every  system  of  equations  in  two 
variables  in  which  one  equation  is  linear  and  the  other  quadratic 
can  he  solved  by  the  method  of  substitution, 

EXAMPLE 

Solve  the  system  |  ^'  +  ^'  =  ^'  (1) 

yx-y  =  l.  (2) 

Solution  :  Solving  (2)  for  x  in  terms  oi  y,     x  =  1  +  y.  (3) 

Substituting  1  +  y  for  x  in  (1),  (1  +  yY  +  y'^  =  ^.  (4) 

From  (4),  /  +  ?/  -  2  =  0.  (5) 

Solving  (5),  3/  =  1  or  —  2. 

Substituting  1  for  y  in  (3),  x  =  1  +  1  =  2. 

Substituting  -2iov  y  in  (3),  a:  =  l-2  =  -l. 

The  two  sets  of  roots  are  x  =  2,  y  =  1  and  x  =  —  1,  ?/  =  —  2. 

Check :  Substituting  2  for  x  and  1  for  y  inj^^^      2  -  1  -  l' 
Substituting  —  1  for  x  and  —  2  for  y  in      \  ^^{^  o  Z  i ' 

LC  /'  —  1  +  2  —  1. 

143 


144  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Solve  the  following  systems,  pair  results,  and  check  each 
Sf^  of  roots : 


r; 


x'  +  y^^  20. 
4,m  +  n  =  28, 
2m^  +  S7nn  =  98. 


m-27iV5  =  0. 


4:S   +   t=6, 

st  =  -  10. 


10. 


3/?j  +  4/?,=:5, 

2R^R^-6R^  =  -3. 

2x2j  +  y^-20  =  0, 

xi/  +  AO  =  0. 

7,2  4.  7,2  ^2  A;  =  40, 

7i  +  7c  +  2  =  0. 

m^  +  3  m?i  +  71^  =  88, 

2  m  =  71. 

0,2  +  2^2  +  4^^6^^40^ 

a;  -  10  =  ?/. 

y  +  ^Vl5  =  0, 

7/2  +  a;«  =  16:r. 

4s-y  =  30. 
a;V3  +  5y  =  -72, 
^•xy  =  -15V3.  ^^- 

If  the  equations  of  a  system  are  not  one  linear  and  the  other 
quadratic,  an  attempt  to  solve  it  by  substitution  usually  gives 
an  equation  of  the  third  or  fourth  degree  at  least.  In  most 
cases  such  an  equation  could  not  be  solved  by  factoring,  and  at 
the  present  time  its  solution  by  any  other  method  is  beyond 
the  student.  The  various  devices  explained  in  the  following 
pages  are  for  the  purpose  of  avoiding  the  necessity  of  solving 
an  equation  of  a  higher  degree  than  the  second. 

77.  Homogeneous  equations.  An  equation  is  homogeneous  if,  on 
Ijeing  written  so  that  one  member  is  zero,  the  terms  in  the  other 
member  are  of  the  same  degree  with  respect  to  the  variables. 

[        Thus  x^  -\-  y^  =  xy  and  x^  —  Sxy  -\-  y^  =  0  are  homogeneous 
\  equations  of  the  second  degree ;   2x^  -^  y^  =  xhj  —  3  xy'^  is  a 
homogeneous  equation  of  the  third  degree. 

78.  Both  equations  quadratic.  If  the  system  is  of  either  type  de- 
scribed in  the  following  examples,  it  can  be  solved  by  quadratics. 

The  first  example  illustrates  the  type  when  one  equation, 
but  not  necessarily  both  of  them,  is  homogeneous. 


SYSTEMS  SOLVABLE  BY  QUADRATICS  145 

EXAMPLES 

011  .       (Sx^  +  4:f  =  Sxi/,  (1) 

1.  Solve  the  system  i    ,   ,     .  %.         o  /o\ 

[  ij^  +  x^  —  o  X  =  S.  (J) 

Solution :  First  we  solve  the  homogeneous  equation  (1)  for  x  in 
terms  of  y. 

Transposing  in  (1),   ^  x'^  -  S  xy  +  i  y"^  =  0.  (3) 

Solving  (3)  by  formula,  x  =  «  y  ±  V64y^- 48y^     ^^^ 

Whence  •     x  =  2  y  or  -^ .  y    (^) 

Substituting  2  y  for  x  in  (2), 


y 


2/2  +  4  ?/2  _  10  ?/  =  3.  X       (6) 

Solving  (6),  y  =  l±|VlO.    \y         (7) 

By  (5),  x  =  2y]  then  from  (7),  x  =  2±\  VlO.  •         (8) 

2  V 

Substituting  -^  for  x  in  (2), 

,.  +  i|!-Hl  =  3.  (9) 

Solving  (9),  2/  =  3  or  -  ^%.  ^        (10) 

By  (5),  x  =  y^\  then  from  (10),        x  =  2  or  -  ^-^.  \     (11) 

3 

Pairing  results, 
x  =  2^^       x  =  -^%^  x  =  2  +  tVl0l^       :.3.2-|Vl^l^ 

^  =  :3j^'     y  =  -i%J     '      ?/  =  l  +  fVlOj     '      7/  =  l-fVlOj 
Check : 

r  3  (2)2  +  4  (3)2  =  8  .  2  .  3,  or  48  =  48. 
132  +  22-5-2  =  3,  or  3 -3. 

^  r  3  (-  ^V)^  +  4  (-  ^V)'  =  8  •  (-  tV)  (-  tV)'  or  fit  =  f  f  f . 
U-  1%)^  +  (-  A)'  -  5  •  (  -t\)  =  3,  or  3  =  3. 

c  r  3  (2  ±  4  vTo) 2  +  4  (1  ±  I  VTo)'  =  8  (2  ±  f  Vio)  (1  ±  f  Vio) . 
^  1  (1  ±  I  Vio)'  +  (2  ±  I  VTo)'  -  5  (2  ±  I  Vio)  =  3. 

Taking  both  values  in  C  and  D  with  the  sign  +  or  both  with 
the  sign  — , 

r  12  ±  -\8.  vio  +  -\6  +  4  ±  1/  vio  +  -\2  =  16  ±  -^vio  +  i|^. 

|l  ±  I  VlO  +  f  +  4  ±  -1/  VTo  +  -\2  _  10  T  4  VlO  =  3. 

If  each  equation  of  a  system  in  two  variables  is  quadratic  and 
both  are  homogeneous  with  the  exception  of  a  constant  term  (not 
zero),  the  system  is  solved  much  like  the  preceding  one. 


146        SECOND  COURSE  IN  ALGEBRA 

[x^  +  y^  =  10.  (2) 

Hint.   First  we  combine  the  two  equations  to  obtain  a  homo- 
geneous equation  in  which  the  constant  term  is  zero. 

(1).5,  5x^  +  15/ =  30.  (3) 

(2)  .3,  3  a:2  +  3  r/2  =  30.  (4) 

(3)  -  (4),  -  3  a:2  +  5  a:?/  +  12  y2  ^  0.  (5) 

Solving  (5)  for  x  in  terms  of  y,  x  =  3  ?/  or f^-  (6) 

3 

We  can  now  substitute  from  (6)  in  (2)  and  proceed  precisely  as  in 

the  last  example.  The  student  should  complete  the  work  and  obtain 

a;  =  3,  -3,  +  |  VTo ,  -  |  VTo. 

y  =  l,         -1,  -f^^,  +fVlO. 


4 


J 


EXERCISES 

Solve,  pair  results,  and  check  each  set  of  real  roots  : 

x'-[-xy  =  ^,  x^-{-X2j-^y^  =  4, 

^-  f-xy  =  10.  x^-2xy  =  12. 

x2-f/  =  10,  x^^xy^2f, 

82/^  +  a-?/  =  6.  2^2  +  a;  =  2  +  7/2. 

u^-\-2uv  =  ^,  x'  +  2xy-y''  =  ^2, 

2v2-f37^v=-16.  2£c2-3a!?/4-2/'=0. 

^2-352^  =  4,  2x'~xy  +  2if  =  12, 

32^  +  35^  =  12.  2x''-\-xy-\-2f  =  S. 

g    a;  (a; +  2?/)  =  16,  ^^    a;^  -  a-?/ -  5 2/^  =  15, 

•   y(y-x)  =  S.  '   a;2-67/2=l. 

Up  to  this  point  the  systems  considered  have  been  solved  by  a 
method  partially  described  by  the  word  "  substitution."  The  essential 
step  in  this  method  is  to  solve  one  of  the  original  equations  (or  one 
derived  from  the  original  system)  for  one  variable  in  terms  of  the 
other,  and  substitute  the  value  found  in  the  other  equation  (or  in 
either  of  the  original  equations).  This  method  is  applicable  more 
frequently  than  those  which  are  given  later.  Consequently  it  is 
much  more  important  for  the  student  to  master  the  method  of  sub- 
stitution than  it  is  for  him  to  master  any  other  method. 


SYSTEMS  SOLVABLE  BY  QUADRATICS 


147 


A 


79.  Equivalent  systems.   Equivalent  systems  of  equations  are 
systems  which  have  the  same  set  or  sets  of  roots. 

The  graphs  of  equivalent  systems  have  common  points  of 
intersection. 

If  we  solve  graphically  the  two  systems 

I x^- 2/^  =  12,(1)  <x-y  =  2,(3) 

lx  +  2/  =  6.       (2)     ""'"^     ^U  +  2,  =  6.  (4) 

we  obtain  the  graphs  of  the  adjacent  figure.  The  hyperbola 
(1)  and  the  straight  line  (2)  intersect  at  only  one  point  (4,  2). 
The  straight  lines  (2)  and 
(3)  intersect  at  this  very 
point.  Hence  the  systems 
A  and  B  are  equivalent. 

80.  Special  devices.  Sys- 
tems of  equations  are 
often  met  which  can  be 
solved  by  substitution, 
but  which  are  more  con- 
veniently solved  as  in  the 
following  illustrations.  It  should  be  observed  that  in  every 
case  the  aim  of  the  device  is  to  replace  the  given  system  by 
an  equivalent  system  of  linear  equations,  or  by  a  system  in 
which  one  equation  is  quadratic  and  the  other  linear. 


V 

^^               ^ 

s            ^.          /. 

^')            ^       zz 

V            s  ^^ 

N                                n                   \^ 

\                           ^                   /^\ 

±_     --I     __'   ^ZI    ^^--.x- 

^'                         /                0/2      1           ^           ^' 

1                 /             \           2K 

7Z     ^          \        "" 

J^   ^            ^^ 

/        3/                              (l\ 

z     z                  ^^ 

/        '                               ^ 

"^                  y' 

EXAMPLES 


1,  Solve  the  system 

Solution :  Squaring  (1), 
(2) -4, 
(3) -(4), 
From  (5), 


xy  =  6. 


7, 

x'^  +  2  xij  +  if  =  49. 
4zxy  =  24. 
x^  —  2  xy  +  y^  =  25. 
X  —  y  =±  5, 


From  (6)  and  (1),  ^|^"^^ 
For  A,  X  =  Q,  y 


7,  (1) 
lx-y  =  6.  (7) 
1 ;  and  for  B,  x  =  1, 


Bi 


Cx  +  y  =  7, 


IX 
=  6. 


(1) 
(8) 


(1) 

(2) 

(3) 
(4) 
(5) 
(6) 


us 


SECOND  COURSE  IN  ALGEBRA 


The  derived  systems  -.4  and  B  are  equivalent  to  the  original  sys- 
tem (1),  (2).  The  graphs  of  the  adjacent  figure  show  that  the  straight 


^ 

r — 

n 

\ 

1     , 

• 

\i/ 

(6) 

/ 

K 

/ 

\ 

\ 

/ 

\ 

\ 

/ 

\ 

\ 

f7). 

• 

/ 

V 

^v 

/ 

(2) 

J 

' 

/ 

"n 

X- 

,/ 

0 

'. 

I   : 

/ 

\ 

/ 

^ 

/ 

(1) 

/ 

(2N 

/ 

\ 

• 

\ 

/ 

/' 

, 

/ 

1 

y' 

line  (1)  and  the  hyperbola  (2)  have  the  same  points  of  intersection 
as  the  three  straight  lines  (1),  (7),  and  (8)  of  systems  A  and  B. 

A  method  similar  to  that  of  the  preceding  solution  can  be  applied 
to  the  following  system  : 


2.   Solve 


x'  +  f  =  37, 


[x'  +  f 
{  xy  =  6. 


Solution:  (2) -2, 

(1)  +  (3), 
From  (4), 

a)-(3), 

From  (6), 


2xy  =  12. 
x^ -\- 2  zy  ■{■  y^  =  49. 

x  +  y=±7. 
x^-2xy  +  y'^  =  25. 

X  —  y  =±  6. 


(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 


(5)  and  (7)  combined  give  four  systems  of  equations : 

(8)  fx  +  y  =  -7,    (11) 

(9)  '^\x-y  =  5,  (9) 

,^x  +  y  =  7,  (8)  j^fx  +  y  =  -7,   (11) 

IX-  y  =  -5. 


(x^y  =  7, 
\x-y  =  b. 


B 


(10) 


\^x  —  y  =  —  0 


(10) 


The  solution  oi  A,  B,  C,  and  D  is  left  to  the  student. 
In  the  figure  on  page  149  the  graphs  of  (1)  and  (2)  are  the  circle 
and  the  hyperbola  respectively,  the  two  curves  having  four  points  of 


SYSTEMS  SOLVABLE  BY  QUADRATICS 


149 


intersection.  The  graphs  of  the  four  equations  in  the  systems  A, 
B,  C,  and  D  are  the  four  straight  lines.  These  four  straight  lines 
intersect  in  the  four  points  in  which  the  hyperbola  and  the  circle 


Y 

\ 

^y  / 

.^^          /    ^"^^ 

^         ''    V^   \ 

-T     /   3-X^  V 

A      /            S    nV 

^    /                ^-S-' 

*•-          L.                                       ^S      X 

"         '4'-                             0        2     3./              V    >• 

''^"^                  Z    H 

A^-S           Z      1 

^     ^     \          /            ' 

"^^'^^  /^  ^^ 

/'K 

1       \ 

y' 

intersect.  This  shows  that  the  four  sets  of  roots  belonging  to  the 
system  (1),  (2)  are  identical  with  the  four  sets  belonging  to  the 
four  systems  A,  B,  C,  and  D;  that  is,  the  one  system,  (1)  (2),  is 
equivalent  to  the  four  systems  A,  B,  C,  and  D. 


EXERCISES 


Solve   in  a  manner   similar  to  that  of  the  two  preceding 
examples,  pair  results,  and  check  each  set  of  real  roots  : 


xi/  =  5. 

2. 

x  +  27/  =  S, 

xi/  +  6  =  0. 

3. 

a;2 -1-4  7/ =  101, 

xj/  +  5  =  0. 

4. 

6x-y  =  24., 

360^2  +  7/2  =  288 

4.x' 
xy- 


20  =  0. 


6. 


4cc2_|.y2^25, 

4  a?^  +  4  XT/  +  7/2  =  49. 


0^2^47/2  =  15, 
ic  +  27/  =  3V3. 

g    x'-2xy  =  l^, 
'  2  7/2-xy  =  -6. 


150  SECOND  COURSE  IN  ALGEBRA 


9. 


1     1    .0 

3  ic  -  3  2/  =  7, 

x^  +  P  =  l^' 

11. 

9x2  +  18cc?/  +  92/'  = 

1 

i  =  6. 
xy 

12. 

^+J  =  ^''' 

x^      xy^f 

=  7, 

X      y 

i-i  =  l. 

13. 

a;^  +  4/  =  c, 

x      y 

4  ic?/  =  c?. 

10. 


81.  Use  of  division  in  equations.  Sometimes  an  equation  sim- 
pler than  either  of  those  given  can  be  derived  from  a  system 
by  dividing  the  left  and  right  members  of  the  first  equation  by 
the  corresponding  members  of  the  second.  Then  the  equation 
so  obtained  taken  with  one  of  the  first  two  gives  a  derived 
system  more  simple  than  the  original  one  but  not  always 
equivalent  to  it.  The  conditions  under  which  the  two  are 
equivalent,  however,  is  easily  stated  and  explained. 

Theorem.  Let  U,  V,  K,  and  R  he  rational  integral  expres- 
sions in  two  unknowns,  x  and  y.    Then  the  system 

UK  =  VM,  (1) 

U  =  V  ^  (2) 

is  equivalent  to  the  two  systems 

^=^'  (^)      and      ^=^'  ^*) 

U  =  V,         (2)  V=0.         (5) 

Proof.  Substituting  U  for  V  in  (1),  transposing,  and  factoring,  gives 

U{K-R)  =  0.  (6) 

From  (2),  U-V=0.  (7) 

But  the  system  (6),  (7)  is  equivalent  to  the  two  systems  (3),  (2) 
and  (4),  (5).  This  is  at  once  apparent  since  it  can  be  seen  from 
inspection  that  any  set  of  roots  which  satisfies  (3),  (2)  or  (4),  (5) 
will  satisfy  (6),  (7);  and  conversely. 

Now  if  C7  or  F  is  an  arithmetical  number  (not  zero),  the  system 
(3),  (2)  alone  is  equivalent  to  the  original  one,  since  either  (4) 
or    (5)    would   not  in  that  case  involve  any  unknown. 


SYSTEMS  SOLVABLE  BY  QUADRATICS  151 

Therefore  in  a  system  of  the  form  of  (1),  (2)  we  may  use 
division  and  thereby  obtain  one  simpler  equivalent  system  if 
U  or  V  is  an  arithmetical  number.  In  any  other  case  we  can 
at  once  write  down  the  two  systems  which  are  equivalent  to 
the  original  one.  Either  of  these  courses  makes  it  easier  to 
obtain  all  the  sets  of  roots  which  satisfy  the  original  system. 

EXAMPLES 

In  Examples  1,  2,  and  3  division  gives  in  each  case  the  one 
equivalent  system  on  the  right. 

x^  —  y-  =  12,  X  —  y  =  2,      (See  graph,  p.  147.) 

'  X  -\-  y  =  Q>.  X  -{-y  =  Q.      (One  set  of  roots.) 

^    x''-y^  =  4.x  +  Qy-^,    x  +  y  =  2x  +  ^y -4., 
'  X  —  y  =  2.  X  —  y  =  2.     (One  set  of  roots.) 

x^  +  y^  =  28,  x^-xy-{-y''  =  7.  (See  graph,  p.  152.) 

'  cc  +  ?/  =  4.  a;  +  2/  =  4.  (Two  sets  of  roots.) 

a,3_    8^g^_^3         JL8,     ^,.   .  .        .       ^,     ^ 
4.  „  „  Division  gives  the  two  systems : 

x-y  =  2x  +  y-&.  ^  ^ 

I  x^  +  a:7/  -h  /  =  3,  ^^^^  {x-y  =  0, 

[jj  +  2?/  =  6,  \2x-\-y  —  &  =  {).  (Three  sets  of  roots.) 

The  first  system  in  Example  3  has  two  sets  of  roots,  that  in 
Example  4  has  three.  Hence  the  use  of  division  without  a 
correct  use  of  the  theorem  on  page  150  would  frequently 
result  in  an  incomplete  solution.  If  time  permits^  the  student 
should  graph  the  equations  of  Example  4. 

EXERCISES 

Solve  (using  division  where  possible),  pair  results,  and  check 
each  set  of  real  roots  : 


4^2  _  ^2  ^^g^ 

1        1       if; 

2x  +  y  =  S. 

i  +  Ua 

X      y 

RVi  -  75  =  0, 
Rh  =  15. 

152 


SECOND  COURSE  IN  ALGEBRA 


10 


9  h'^k- 100  =  0, 
3  k^h  +  80  =  0. 
P(l  +  r)2  =  224.72, 
P  +  Pr  =  212. 
9  x^if  +  6  =  15  xy, 
3a;?/ 4- 4  =  6. 
a:*  =92/* +  48, 
ir2  =  3/  +  2. 

|-' =  .16,  ^^  =  3.2. 

9.  1  —  X  =  y,  1  —  x^  =  y^. 
a;2- 2  a;?/ -24  2/2  =  32, 
X  —  6y  =  2. 


11. 


12. 


13. 


14. 


15. 


16. 


4:X^  +  7=Sxy-{-5y% 
4a;2  =  l+  2/2. 

cc«  +  2/8  =  4a;-62/-8, 
X  -\-  y  =  2x  —  Sy  —  \. 

x^  —  y^  =  6x, 
X  —  y  =  Sx. 

x'  +  f=2S, 
a;  +  2/  =  4. 

x^-xy-{-y^  =  7, 
x^-hy'=  28. 
X  +  2/  =  4,  ■ 
x^  —  xy  -\-  y'^  =  7. 


In  the  following  figure  I,  II,  and  III  are  the  graphs  of  x^  -^  y^  =  28, 
x'^  —  xy  +  y^  =  7,  and  x  -\-  y  =  4:  respectively.  These  equations  are 
taken  from  the  systems  in  Exercises  14,  15,  and  16  which  contain 


but  three  different  equations  paired  in  three  ways.  Since  the  two 
sets  of  roots  for  each  is  the  same,  we  know  that  the  three  systems 
are  equivalent.  The  equivalence  of  the  three  systems  is  also  shown 
in  the  preceding  figure. 


SYSTEMS  SOLVABLE  BY  QUADRATICS  153. 

MISCELLANEOUS    EXERCISES 

Solve  by  any  method,  pair  results,  and  check  each  set  of 
real  roots : 

(If  any  system  cannot  be  solved  algebraically  by  the  methods 
previously  given,  solve  it  graphically.) 

2x^  +  /  =  33, 
x^-\-2if  =  54. 


2. 

3A-^-8/i;'^  =  40, 

5  A2  +  /.2  ^  gi_ 

3. 

4  Rl  +  3  --  9  111, 

12  Rl  +  72|  =  V. 

4 

xy  -\-x  =  18, 

*• 

X7J  +  7J  =  20. 

5. 

ic-' 

=  y^ 

xy 

=  8. 

6 

X  - 

-xy  = 

=  5, 

u. 

2y  +  xy 

=  6. 

7. 

x' 

-f  = 

:19, 

X  - 

-y  = 

1. 

8. 

x' 

-f  = 

:19, 

x^ 

+  xy  +  y' : 

19. 

g    x''  +  xy  +  y''  =  l'd, 
'  x-y  =  l. 
10.  Show  that  the  systems  (7),  (8),  and  (9)  are  equivalent  by 
graphing  the  three  equations  of  these  exercises. 

^s^-2f  =  (},  x^  +  xy  +  x  =  (i, 

^s'-^f^l,  x^  +  xy  +  2x  =  0. 

12.  9  19.    ,  ;    -^     -^      ' 

2n'-^  =  m\  x''-\-xy  +  x  =  0. 

5lFi^- 6.811^1=99.55,  1+1^13 

^'*-   1^1^-171  =  20.  20.  ^ 

*  x^  +  2y^  =  2,  ^-^  =  1- 

a^2^_2^2^  +  22/^  =  10,  i-i  =  7 

^^'  3  ic-2  -xy-^f  =  51.  „,    a^'      2/'       •' 

i_i  =  l 
x      y 

x"  -2xy+2if-y=0, 
^^'  2x2-3:r7/-/  +  22/  =  0 


^^-  a.^+2/  =  ll. 

0.^  +  2/^  =  10. 

=  7, 

24 
25 


154  SECOND  COURSE  IN  ALGEBRA 

x^  ^z^  =  34, 

23.  a;2  +  2/'  =  25, 

f^z^^  41. 

3icy  =  a;y-88, 
a;  —  2/  =  6. 

a^«  =  2/8  +  37, 

■i-15  +  l  =  9 

ic^       xy       y^ 
JL_i_3 

xy       f 

a;2  =  8£c  +  62/, 
2/2  =  6a;4-82/. 

x8-2/«  =  2a^  +  2/-4, 
ic  4-  2  ?/  =  4. 


27. 
28. 
29. 


32. 

x-2y-2      4 
1      1_  1 

X      y~  12' 

33. 

£c2  +  6ic-2  =  36^. 

3:r2/-a;2  =  36. 

34. 

iP  —  2/  V^  =  24, 

a;2  +  xf  =  320. 

35. 

x-1 

y'  +  y-hl      13 

a;2  _  a;  +  1      43 

36. 

^.  +  ^  =  16, 
X        xy 

-5 1 — 5=0. 

x^      xy      f 

xy  =  c, 

X  +  y  =  a.  Z7.  9  X  ^  y  =  IS  =  xy. 


30.  1,1      o'  38.  4a?  +  -  =  46=-z 

x-^  +  y-^  =  2.  y  5        y/ 

31.  V  ^  r  ^^'  39.  ^'  +  2/'  -  (y  -  ^)  =  12, 
ic*  —  y*  =  2.  '  x^  —  xy  =  0. 


PROBLEMS 

(Reject  all  results  which  do  not  satisfy  the  conditions  of  the 
problems.) 

1.  Find  two  numbers  whose  difference  is  4  and  the  differ- 
ence of  whose  squares  is  88. 

2.  The  sum  of  two  numbers  is  21  and  the  sum  of  their 
squares  is  189.    Find  the  numbers. 

3.  Find  two  numbers  whose  product  is  192  and  whose  quo- 
tient is  |. 


SYSTEMS  SOLVABLE  BY  QUADRATICS  155 

4.  The  area  of  a  right  triangle  is  150  square  feet  and  its 
hypotenuse  is  25  feet.    Find  the  legs. 

5.  A  rectangular  field  is  8  rods  longer  than  it  is  wide  and 
the  area  of  the  field  is  8  acres.   Find  the  length  and  the  width. 

6.  The  difference  of  the  areas  of  two  squares  is  252  square 
feet,  and  the  difference  of  their  perimeters  is  24  feet.  Find  a 
side  of  each  square. 

7.  The  area  of  a  rectangular  field  is  3f  acres  and  one  diago- 
nal is  60  rods.    Find  the  perimeter  of  the  field. 

8.  The  perimeter  of  a  rectangle  is  112  feet  and  its  area  is 
768  square  feet.    Find  the  length  and  the  width. 

9.  A  mean  proportional  between  two  numbers  is  2  VT4,  and 
the  sum  of  their  squares  is  113.    Find  the  numbers. 

10.  The  value  of  a  certain  fraction  is  §.  If  the  fraction  is 
squared  and  44  is  subtracted  from  both  the  numerator  and 
the  denominator  of  this  result,  the  value  of  the  fraction  thus 
formed  is  ■^^.    Find  the  original  fraction. 

11.  The  base  of  a  triangle  is  6  inches  longer  than  its  alti- 
tude, and  the  area  is  |  square  feet.  Find  the  base  and  altitude 
of  the  triangle. 

12.  The  volumes  of  two  cubes  differ  by  1413  cubic  inches 
and  their  edges  differ  by  3  inches.    Find  the  edge  of  each. 

13.  The  sum  of  the  radii  of  two  circles  is  25  inches  and 
the  difference  of  their  areas  is  125  it  square  inches.  Find  the 
radii. 

14.  The  perimeter  of  a  rectangle  is  5  C  and  its  area  is  C^. 
Find  its  dimensions. 

15.  The  area  of  a  right  triangle  is  8  «,^  —  8  ^^  and  its  hy- 
potenuse is  4  V2  a^  +  2  V^.    Find  the  legs. 

16.  The  perimeter  of  a  right  triangle  is  ^Q  feet  and  its  area 
is  84  square  feet.    Find  the  legs  and  the  hypotenuse. 


156  SECOND  COURSE  IN  ALGEBRA 

17.  If  a  2-digit  number  be  multiplied  by  the  sum  of  its 
digits,  the  product  is  324;  and  if  three  times  the  sum  of  its 
digits  be  added  to  the  number,  the  result  is  expressed  by 
the  digits  in  reverse  order.    Find  the  number. 

18.  The  yearly  interest  on  a  certain  sum  of  money  is  |42. 
If  the  sum  were  |200  more  and  the  interest  1%  less,  the 
annual  income  would  be  |6  more.  Find  the  principal  and  the 
rate. 

19.  A  wheelman  leaves  A  and  travels  north.  At  the  same 
time  a  second  wheelman  leaves  a  point  3  miles  east  of  A  and 
travels  east.  One  and  one-third  hours  after  starting,  the 
shortest  distance  between  them  is  17  miles ;  and  3^  hours 
later  the  distance  is  53  miles.    Find  the  rate  of  each. 

20.  The  circumference  of  the  fore  wheel  of  a  carriage  is 
1  foot  less  and  that  of  its  rear  wheel  3  feet  less  than  the 
circumferences  of  the  corresponding  wheels  of  a  farm  wagon. 
In  going  1  mile  the  fore  wheel  of  the  carriage  makes  40  revo- 
lutions more  than  its  rear  wheel,  and  the  fore  wheel  of  the 
wagon  makes  88  more  than  its  rear  wheel.  Find  the  circum- 
ferences of  the  carriage  wheels. 

21.  A  starts  out  from  P  to  Q  at  the  same  time  B  leaves  Q 
for  P.  When  they  meet,  A  has  gone  40  miles  more  than  B. 
A  then  finishes  the  journey  to  Q  in  2  hours  and  B  the  journey 
to  P  in  8  hours.  Find  the  rates  of  A  and  B,  and  the  distance 
from  P  to  Q. 

22.  A  leaves  P  going  to  Q  at  the  same  time  that  B  leaves 
Q  on  his  way  to  P.  From  the  time  the  two  meet,  it  requires 
6§  hours  for  A  to  reach  Q,  and  15  hours  for  B  to  reach  P. 
Find  the  rate  of  each,  if  the  distance  from  P  to  Q  is  300  miles. 

23.  A  man  has  a  rectangular  plot  of  ground  whose  area  is 
1250  square  feet.  Its  length  is  twice  its  breadth.  He  wishes 
to  divide  the  plot  into  a  rectangular  flower  bed,  surrounded  by 
a  path  of  uniform  breadth,  so  that  the  bed  and  the  path  may 
have  equal  areas.    Find  the  width  of  the  path. 


SYSTEMS  SOLVABLE  BY  QUADRATICS 


157 


GEOMETRICAL  PROBLEMS 


Find  the  alti- 


1.  The  sides  of  a  triangle  are  6,  8,  and  10. 
tude  on  the  side  10. 

Hint.  From  the  adjacent  figure  we 
easily  obtain  the  system  : 

\  (10  -  x)2  +  ?/2  =  64. 

2.  The  sides  of  a  triangle  are  8, 
15,  and  17.  Find  the  altitude  on 
the  side  17  and  the  area  of  the 
triangle. 

3.  The  sides  of  a  triangle  are  13,  20,  and  21.    Find  the  alti- 
r  tude  on  the  side  20  and  the  area  of  the  triangle. 

4.  The  sides  of  a  triangle  are  7,  15,  and  20.  Find  the  alti- 
tude on  the  side  7  and  the  area  of  the  triangle. 

5.  The  sides  of  a  triangle  are  10,  17,  and  21.  Find  the 
altitude  on  the  side  10  and  the  area  of  the  triangle. 

6.  Find  correct  to  two  decimals  the  altitude  on  the  side  16 
of  a  triangle  whose  sides  are  12,  16,  and  18  respectively. 

7.  The  parallel  sides  of  a  trapezoid  are  14  and  26  respec- 
tively, and  the  two  nonparallel  sides  are  10  each.  Find  the 
altitude  of  the  trapezoid. 

Hint.  Let  A  BCD  be  the  trapezoid.  Draw  CE  parallel  to  DA  and 
CF  perpendicular  to  AB. 

Then  EC  =  10,  AE=U,  Z) ii jj 

and  EB  =  26-  14,  or  12. 
If  we  let  EF  =  x,  FB 
must  equal  12  —  a: ;  then 
we  can  obtain  the  system 
of  equations : 

ra;2  +  2/2^100, 
l(12-a:)2  +  3/2  =  ioo. 

8.  The  two  nonparallel  sides  of  a  trapezoid  are  12  and  17 
respectively,  and  the  two  bases  are  5  and  13  respectively. 
Find  the  altitude  of  the  trapezoid. 


158  SECOND  COURSE  IN  ALGEBRA 

9.  The  bases  of  a  trapezoid  are  15  and  20  respectively,  and 
the  two  nonparallel  sides  are  29  and  30.  Find  the  altitude  of 
the  trapezoid  and  the  area. 

10.  The  sides  of  a  trapezoid  are  12,  20,  17,  and  45.  The 
sides  20  and  45  are  the  bases.   Find  the  altitude  and  the  area. 

11.  The  sides  of  a  trapezoid  are  21,  27,  40,  and  30.  The 
sides  21  and  40  are  parallel.  Find  the  altitude  and  the  area 
of  the  trapezoid. 

12.  The  sides  of  a  trapezoid  are  23,  85,  100,  and  x.  The 
sides  23  and  100  are  the  bases,  and  each  is  perpendicular  to 
the  side  x.   Find  x  and  the  area  of  the  trapezoid. 

13.  The  parallel  sides  of  a  trapezoid  are  42  and  250.  The 
other  sides  are  123  and  325.  Find  the  altitude  and  the  area 
of  the  trapezoid. 

14.  The  area  of  a  triangle  is  1  square  foot.  The  altitude 
on  the  first  side  is  16  inches.  The  second  side  is  14  inches 
longer  than  the  third.    Find  the  three  sides. 


CHAPTER  XII 
PROGRESSIONS 

82.  Definitions.  In  all  fields  of  mathematics  we  frequently 
t  iicounter  groups  of  three  or  more  numbers,  selected  according 
to  some  law  and  arranged  in  a  definite  order,  whose  relations 
to  each  other  and  to  other  numbers  we  wish  to  study. 

The  individual  numbers  or  expressions  are  called  terms. 

In  the  following  examples  the  law  of  formation  and  the 
order  of  the  terms  are  so  obvious  that  the  student  can  write 
down  many  additional  terms. 

EXERCISES 

Write  three  more  terms  in  each  of  the  following : 

5.  1\  3^  52,  .... 

6.  a/I,  ^2,  ^3,.... 

7.  2,4,8,16,.... 

1  1 


1. 

1,2,3,4,5,.... 

2. 

2,4,6,8,.... 

3. 

9,8,7,6,.... 

4. 

-1,-3,-5,- 

^, 

.    .    1 

1 

9.  1, 


1    12    123  .12.3.4         * 

10.  1,  I4-V2,  1  +  2  V2,  .... 

There  is  an  unlimited  variety  of  such  groups  or  successions 
of  numbers.    Only  two  simple  types  will  be  considered  here. 

83.  Arithmetical  progression.    An  arithmetical  progression  is  a 

succession  of  terms  in  which  each  term  after  the  first,  minus 
the  preceding  one  gives  the  same  number. 

This  same  number  is  called  the  common  difference  and  may 
be  any  positive  or  negative  number. 

159 


160  SECOND  COURSE  IN,  ALGEBRA 

The  numbers  3,  7, 11, 15,  •  •  •  form  an  arithmetical  progression, 
since  any  term  after  the  first,  minus  the  preceding  one  gives  4. 
Similarly,  12,  6,  0,  —  6,  —  12,  •  •  •  is  an  arithmetical  progression, 
since  any  term  minus  the  .preceding  one  gives  the  common 
difference  —  6.  In  like  manner,  |,  5,  6^,  •  •  •  is  an  arithmetical 
progression  whose  common  difference  is  1^.. 

EXERCISES 

From  the  following  select  the  arithmetical  progressions,  and 
in  each  of  them  find  the  common  difference : 

1.  4,i»^o.,%...  6.18,8,-2,.... 

2.  10,  16^,  23,  • . ..  7.  5  a  +  2,  3  a  +  1,  «,  . . .. 

3.  2,  4,  8,  ....  8.  3x-5,2x-\-S,x-7,---. 

4.9,12^,15,....  9    _L,A,  V3  .   . 

5.  25,21,17,  ...  V3    V3 

V3-2    2(V3-l)      1 

84.  The  last  or  nth  term  of  an  arithmetical  progression.    If  a 

denotes  the  first  term  and  d  the  common  difference,  any  arith- 
metical progression  is  represented  by 

a,  a  -{-  d,  a  -{-  2  d,  a  +  S  d,  a  -\-  4:  d,  etc. 

Here  one  observes  that  the  coefficient  of  d  in  each  term  is  one 
less  than  the  number  of  the  term.  Hence  the  nth.  or  general 
term  is  a  -f-  (?i  —  1)  d.    If  I  denotes  the  nth.  term,  we  have 

l=a  +  (n-l)d.  (A) 

EXERCISES 

1.  Find  the  12th  term  of  the  progression  1,  5,  9,  13,  ... . 

2.  Find  the  23d  term  of  the  progression  — 18,  — 15,  — 12,  ••  • . 

3.  Find  the  15th  term  of  the  progression  13,  7,  1,  —  5,  •  • .. 

4.  Find  the  19th  term  of  the  progression  a,  3 a,  5a,  ••. 

5.  Find  the  7th  and  12th  terms  of  the  progression  ^,  If,  1, . .  • . 


10. 


i 


PROGRESSIONS  161 

6.  Find  the  5th  and  20th  terms  of  the  progression  1,  <x  +  1, 
l  +  2a,  •••. 

7.  Find  the   10th  term  of  the   progression  7  V2,    5  V2, 
3V2,  •••.  ^ 

8.  Find   the    9th   term   of   the    progression j    2, 

3  -  V3  ^ 

9.  Find  the  (n  —  l)st  term  of  the  progression  a,  a  +  d, 
<i  +2d,  ■■■. 

10.  Find  the  (n  —  2)d  term  of  the  progression  a,  a  -{-  d, 
a-\-2d,.... 

11.  Find  the  (n  —  3)d  term  .of  the  progression  Vs  —  1, 
2V5-2,  3(V5-l),  ■••. 

12.  Find  the  rith  term  of  the  progression  -  J >  2  —  -?  •••. 

n       n  n 

13.  The  first  and  second  terms  of  an  arithmetical  progres- 
sion are  h  and  k  respectively.  Find  the  third  term  and  the 
Tith  term. 

14.  The  first  and  third  terms  of  an  arithmetical  progression 
are  h  and  k.    Find  the  Tith  term. 

15.  A  body  falls  16  feet  the  first  second,  48  the  next,  80  the 
next,  and  so  on.  How  far  does  it  fall  during  the  10th  second  ? 
during  the  nth.  second  ? 

85.  Arithmetical  means.  The  arithmetical  means  between  two 
numbers  are  numbers  which  form,  with  the  two  given  ones  as 
the  first  and  the  last  terms,  an  arithmetical  progression. 

The  insertion  of  one  or  more  arithmetical  means  between 
two  given  numbers  is  performed  as  in  the  following : 

Example  :  Insert  three  arithmetical  means  between  5  and  69 

Solution :  I  =  a  ■\-  (n  —  V)d. 

There  will  be  five  terms  in  all. 

Therefore  69  =  .5  +  (5  -  1)  d. 

Solving,  d  =  16. 

The  required  arithmetical  progression  is  5,  21,  37,  53,  69. 


162  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

1.  Insert  the  arithmetical  mean  between  3  and  15. 

2.  Insert  the  arithmetical  mean  between  h  and  4  k. 

3.  Insert  two  arithmetical  means  between  2  and  17. 

4.  Insert  two  arithmetical  means  between  a  and  h. 

5.  Insert  three  arithmetical  means  between  —  4  and  16. 

6.  Insert  three  arithmetical  means  between  m  and  n. 

7.  Insert  six  arithmetical  means  between  3  and  45. 

8.  Insert  nine  arithmetical  means  between  3  and  ^^: 

9.  Insert  four  arithmetical  means  between  —  V2  and  9  V2. 

10.  Insert  five  arithmetical  means  between  7 x  —  Sa  and 

13a;  +  9a.  *  ^ 

5  15  V  5 

11.  Insert  six  arithmetical  means  between j=.  and  ^— r 

2V5  2 

V2 

12.  Insert  two  arithmetical   means   between   — -;= and 

V2(1-2V2).  ^2-1 

13.  What  is  the  arithmetical  mean  between  any  two  numbers  ? 

14.  In  going  a  distance  of  1  mile  an  engine  increased  its 
speed  uniformly  from  20  miles  per  hour  to  30  miles  per  hour. 
What  was  the  mean  or  average  velocity  in  miles  per  hour 
during  that  time  ?    How  long  did  it  require  to  run  the  mile  ? 

15.  The  velocity  of  a  falling  body  increases  uniformly.  At  the 
beginning  of  the  third  second  its  velocity  is  64  feet  per  second, 
and  at  the  end  of  the  third  second  it  is  96  feet  per  second. 
(a)  What  is  its  mean  or  average  velocity  in  feet  per  second  dur- 
ing the  third  second  ?  (b)  How  many  feet  does  it  fall  during 
the  third  second? 

16.  The  velocity  of  a  body  falling  from  rest  is  32  feet  per 
second  at  the  end  of  the  first  second.  What  is  the  mean  or 
average  velocity  in  feet  per  second  during  the  first  second  ? 
How  many  feet  does  the  body  fall  during  the  first  second? 
the  second  second? 


PROGRESSIONS  163 

17.  Find  the  mean  or  average  length  of  25  lines  whose  lengths 
in  inches  are  the  first  25  even  numbers.  ' 

18.  rind  the  mean  length  of  17  lines  whose  lengths  in  inches 
are  given  by  the  consecutive  odd  numbers  beginning  with  11. 

19.  With  the  conditions  of  Problem  15  determine  the  average 
velocity  per  second  of  a  body  which  has  fallen  for  10  seconds. 

20.  A  certain  distance  is  separated  into  8  intervals,  the 
lengths  of  which  are  in  arithmetical  progression.  If  the  shortest 
interval  is  1  inch  and  the  longest  22  inches,  find  the  others. 

86.  Sum  of  a  series.  The  indicated  sum  of  several  terms  of 
an  arithmetical  progression  is  called  an  arithmetical  series.  The 
formula  for  the  sum  of  n  terms  of  an  arithmetical  series  may 
be  obtained  as  follows  : 

S=a-\-(a  +  d)-{-(a  +  2d)  +  --.  +  (l-2d)-\-(l-d)+L  (1) 
Reversing  the  order  of  the  terms  in  the  second  member  of  (1), 
S=l-\-(l-d)-\-(l-2d)  +  ...  +  (a  +  2d)-\-(a  +  d)-\-a.  (2) 
Adding  (1)  and  (2),  f 

2S  =  (a-\-l)-{-(a-{-l)-{-(a  +  l)-{-...  +  (a+l)  +  (a  +  l)  +  (a  +  l) 
=  n{a  -{-  I). 

Therefore  S  =  ^{a-\-l),  (B) 

it 

Substituting  for  I  from  (.4),  page  160, 

^     ■""•         S  =  5[2a+(«-l)d].  (C) 

^Y     \  EXAMPLE 

Required  the  sum  of  the  integers  from  11  to  99  inclusive, 
Solution  :  ?z  =  89,  a  =  11,  Z  =  99. 


w^     ,   ;.     .         e      89(11  +  99) 
-(«  +  /)  gives  S  =  —^ 1 

Therefore  the  sum  of  the  integers  from  11  to  99  is  4895. 


Substituting  in  (B),  S  =  ^(a  +  l)  gives  S  =  ^^"<^^+  ^^)  =  4^95, 


164  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

1.  Find  the  sum  of  10  terms  of  the  series  2  -}-  5  +  8  H . 

2.  Find  the  sum  of  18  terms  of  the  series  10  +  8  -+-  6  H . 

3.  Find  the  sum  of  10  terms  of  the  arithmetical  progression 
3,4i,V-,  •••• 

4.  Find  the  sum  of  12  terms  of  the  arithmetical  progression 
18,14^,11,.... 

5.  Find  the  sum  of  the  first  one  hundred  integers. 

6.  Find  the  sum  of  the  first  one  hundred  even  numbers. 

7.  Find  the  sum  of  the  first  one  hundred  odd  numbers. 

8.  Find  the  sum  of  the  even  numbers  between  187  and  433. 

9.  Find  the  sum  of  the  first  n  odd  numbers. 

10.  Find  the  sum  of  the  first  n  even  numbers. 

11.  How  many  of  the  natural  numbers  beginning  with  1 
are  required  to  make  their  sum  903  ? 

Hint.    Substitute  in  formula  (C)  preceding. 

12.  How  many  terms  must  constitute  the  series  5  +  9  + 
13  +  •   -in  order  that  it  may  amount  to  275  ? 

13.  Beginning  with  80  in  the  progression  78,  80,  82,  how 
many  terms  are  required  to  give  a  sum  of  510?    Explain. 

14.  The  second  term  of  an  arithmetical  progression  is  —  7 
and  the  seventh  term  is  18.    Find  the  eleventh  term. 

15.  Find  the  sum  of  t  terms  of  the  arithmetical  progression 
1    t-1 

16.  If  Z  =  29,  a  =  2,  and  d  =  3,  find  n  and  5. 

17.  If  a  =  S,  d  =  4:,  and  s  =  300,  find  n  and  I. 
f 

18.  If  d  =  —  11,  n  =  13,  and  5  =  0,  find  a  and  I. 

19.  The  first  and  second  terms  of  an  arithmetical  progres- 
sion are  h  and  k  respectively.  Find  the  sum  of  n  terms  of 
the  progression. 


PROGRESSIONS  165 

20.  Us=-33,  a  =  5x-\-2,  and  n  =  11,  find  I  and  d. 

21.  If  6-  =  40  V2,  a  =  -  5  V2,  and  <^  =  2  V2,  find  n  and  Z. 

22.  A  clock  strikes  the  hours  but  not  the  half  hours.  How 
many  times  does  it  strike  in  a  day  ? 

23.  A  car  running  30  miles  an  hour  is  started  up  an  incline, 
which  decreases  its  velocity  2  feet  a  second,  (a)  In  how  many 
seconds  will  it  stop  ?    (b)  How  far  will  it  go  up  the  incline  ? 

24.  A  car  starts  down  a  grade  and  moves  4  inches  the 
first  second,  12  inches  the  second  second,  20  inches  the  third 
second,  and  so  on.  (a)  How  fast  does  it  move  in  feet  per 
second  at  the  end  of  the  twenty-first  second  ?  (b)  How  far 
has  it  moved  in  the  twenty-one  seconds  ? 

25.  An  elastic  ball  falls  from  a  height  of  20  inches.  On  each 
rebound  it  comes  to  a  point  J  inch  below  the  height  reached  the 
time  before.  How  often  will  it  drop  before  coming  to  rest? 
Find  the  total  distance  through  which  it  has  moved. 

26.  The  digits  of  a  3-digit  number  are  in  arithmetical  pro- 
gression. The  first  digit  is  2  and  the  number  is  17|^  times  the 
sum  of  its  digits.    Find  the  number. 

27.  A  clerk  received  |75  a  month  for  the  first  year  and  a 
yearly  increase  of  $50  for  the  next  ten  years.  Find  his  salary 
for  the  eleventh  year  and  the  total  amount  received. 

28.  Fifty  dollars  was  deposited  in  a  bank  every  first  of 
March  from  February  28,  1893,  to  March  2,  1904.  If  the 
money  drew  simple  interest  at  3%,  find  the  amount  due  the 
depositor  on  March  1,  1905. 

29.  Assuming  that  a  ball  is  not  retarded  by  the  air,  deter- 
mine the  number  of  seconds  it  will  take  to  reach  the  ground  if 
dropped  from  the  top  of  the  Washington  Monument,  which  is 
555  feet  high.    With  what  velocity  will  it  strike  the  ground  ? 

30.  A  ball  thrown  vertically  upward  rose  to  a  height  of 
256  feet.  In  how  many  seconds  did  it  begin  to  fall?  With 
what  velocity  was  it  thrown  ? 


166  SECOND  COURSE  IN  ALGEBRA 

31.  A  ball  thrown  vertically  upward  returned  to  the  ground 
7  seconds  later.  How  high  did  it  rise  ?  With  what  velocity 
was  it  thrown  ? 

32.  A  pyramid  of  billiard  balls  stands  on  an  equilateral 
triangle,  10  balls  on  a  side.  How  many  balls  are  there  in  the 
bottom  layer  ?  in  the  whole  pyramid  ? 

33.  A  and  B  start  from  the  same  place  at  the  same  time 
and  travel  in  the  same  direction,  A  travels  12  miles  daily. 
B  goes  7  miles  the  first  day,  7^  miles  the  second,  8  miles  the 
third,  and  so  on.    When  are  they  together? 

34.  A  leaves  P  and  travels  south  2  miles  the  first  day,  4 
the  second,  6  the  third,  and  so  on.  Five  days  later  B  leaves 
P  and  travels  south  at  the  uniform  rate  of  28  miles  a  day. 
When  are  they  together? 

Note.  In  the  earliest  mathematical  work  known  a  problem  is 
found  which  involves  the  idea  of  an  arithmetical  progression.  In 
the  papyrus  of  the  Egyptian  priest  Ahmes,  who  lived  nearly  two 
thousand  years  before  Christ,  we  read  in  essence,  "  Divide  40  loaves 
among  5  persons  so  that  the  numbers  of  loaves  that  they  receive 
form  an  arithmetical  progression,  and  so  that  the  two  who  receive 
the  least  bread,  together  have  one  seventh  as  much  as  the  others." 
From  that  time  to  this,  the  subject  has  been  a  favorite  one  with 
mathematical  writers,  and  has  been  esrtended  so  widely  that  it 
would  require  several  volumes  to  record  all  of  the  discoveries 
regarding  the  various  kinds  of  series. 

87.  Geometrical  progression.    A  geometrical  progression  is  a 

succession  of  terms  in  which  each  term  after  the  first,  divided 
by  the  preceding  one  always  gives  the  same  number. 

The  constant  quotient  is  called  the  ratio. 

The  numbers  2, 10, 50, 250,  •  •  • ,  form  a  geometrical  progression, 
since  any  term  after  the  first,  divided  by  the  preceding'  one 
gives  the  same  number  5.  Similarly,  the  numbers  3,-3  V2, 
6,-6  -\/2,  ■  •  • ,  form  a  geometrical  progression,  since  any  term 
after  the  first,  divided  by  the  preceding  one  gives  the  common 
ratio  —  V2. 


PROGRESSIONS  167 


EXERCISES 


Determine  which  of  the  following  are  geometrical  progres- 
sions and  find  in  each  case  the  corresponding  ratio : 

1.  2,6,18,....  8.   V^,  V6,  Vsi,.... 

2.  15,  5,  1,  ....  9    —    -—    2 

3.  18,  -3,  ^,  ....  '   Vs'         2    '    '*■'• 

4.  2,  4,  16,  ....  10.  7a,  35a^l75a^.... 

5.  8,  -  4  V2,  4,  . . ..  11.  8  VS,  -  2  V5,  V5_,  . . .. 

6.  V2,  VT,  1  a/2,  . . . .  12.  5 x\  10 x%  20 xhf,  ■■■• 

7.  1,3,9,81,....  13.  30-?/^,  12ic^7/,  48yV••. 
14.  Find  the  condition  under  which  a,  b,  and  c  form  a  geo- 
metrical progression. 

88.  The  nth  term  of  a  geometrical  progression.  If  a  denotes' 
the  first  term  and  r  the  ratio,  any  geometrical  progression  is' 
represented  by  a,  ar,  ar^,  ar^,  •  •  -.  It  is  evident  that  the  ex- 
ponent of  r  in  any  term  is  one  less  than  the  number  of  the 
term.  Therefore  if  t,^  denotes  the  nth  or  general  term  of  any 
geometrical  progression, 

tn  =  ^r"-K  (A) 

EXERCISES 

1.  rind  the  fifth  term  of  4,  12,  36. 
Solution  :  Here  a  =  4,  r  =  3,  n  —  1  =  4. 
Substituting  these  values  in  the  formula  t„  =  ar'^-'^, 

^g  =  4  .  34  =  324. 

2.  Find  the  tenth  term  of  3,  6,  12,  . . .. 

3.  Find  the  eighth  term  of  2,  3,  |,  ■  •  • . 

4.  Find  the  twelfth  term  of  5,  -  10,  20,  ... . 

5.  Find  t^  of  the  geometrical  progression  $100,  |106, 
$112.36,.... 

6,.  Find  t  of  the  geometrical  progression  18,  —  6,  +  2,  . . . . 


168  SECOND  COURSE  IN  ALGEBRA 

97  n 

7.  Find  t^^  of  the  geometrical  progression  12  a,  9  a,  —r-  >  •   • . 

—  2c  —  3 

8.  Find  ^^  of  the  geometrical  progression  — ^  ?  —  1,  — —  >  ••  • . 

9.  Find  t^  of  the  geometrical  progression  4  V2,  4,  2  V2. 

10.  Find  t^  of  the  geometrical  progression -= ,  - ,  — —  • 

3  2  V2 

11.  Find  ^g  of  the  geometrical  progression ^)  1,  — - — 

12.  The  nth.  term  of  a  geometrical  progression  is  «?•"  -  ^  What 
is  the  (n-l)st  term?  the  (7i-2)d?  the  (n-S)d?  the 
(?i  +  l)st  ?   the  (71  +  2)d  ? 

13.  The  first  and  second  terms  of  a  geometrical  progression 
are  h  and  k  respectively.    Find  the  next  two  terms. 

89.  Geometrical  means.  Geometrical  means  between  two  num- 
bers are  numbers  which  form,  with  the  two  given  ones  as  the 
first  and  the  last  terms,  a  geometrical  progression. 


EXERCISES 

1.  Insert  two  geometrical  means  between  9  and  72. 

Solution :  There  are  four  terms  in  the  geometrical  progression, 
=  9,  n  =  4,  and  t„  =  t^  =  72. 
Substituting  these  values  in  t^^  =  ar^  - 1, 

72  =  9  r8. 
Whence  r  =  2. 

The  required  geometrical  progression  is  9,  18,  36,  72. 

2.  Insert  two  geometrical  means  between  6  and  48. 

3.  Insert  three  geometrical  means  between  6  and  486. 

4.  Insert  one  geometrical  mean  between  4  and  9. 

5.  Insert  one  geometrical  mean  between  a^°  and  a^'^. 

6.  Insert  three  geometrical  means  between  — 144  and  —  9. 


PROGRESSIONS 


169 


7.  The  fifth  term  of  a  geometrical  progression  is  32,  the 
ninth  term  is  512.    Find  the  eleventh  term. 

8.  The  second  term  of  a  geometrical  progression  is  3  V^, 
the  fifth  term  is  y\.    Find  the  first  term  and  the  ratio. 

9.  Show  that  the  geometrical  means  between  h  and  k  are 

±Vm. 

10.  The  first  and  fourth  terms  of  a  geometrical  progression 
are  h  and  k.    Find  the  second  and  third  terms. 

11.  Insert  three  geometrical  means  between  a  and  c. 

12.  The  sum  of  the  first  and  third  terms  of  a  geometrical 
})rogression  is  13  and  the  second  term  is  6.    Find  each  term. 

13.  In  the  adjacent  figure  ABC  is 
a  right  triangle  and  AD  is  perpendic- 
ular to  the  hypotenuse  BC.  Under 
these  conditions  the  length  of  AD 
is  ahuays  a  geometric  mean  between 
the  lengths  of  BD  and  DC. 

{a)  liBD  =  4.  and  DC  =  9,  find  AD. 

(b)  It  BC  =  26  and  AD  =  12,  find  BD  and  DC. 

14.  In  the  adjacent  figure  AB 
touches  and  AD  cuts  the  circle. 
Under  such  conditions  the 
length  of  AB  is  always  a  geo- 
metric mean  between  the  lengths 
of  AC  and  AD. 

(a)  If  ^Z>=16 
and  AC  =  9,fLndAB. 

(b)  If  DC  =  24  and  yl5  =  16,  find  A  C  and  AD. 

90.  Geometrical  series.  Let  S^  denote  the  indicated  sum  of  n 
terms  of  a  geometrical  progression.  This  indicated  sum  is 
called  a  geometrical  series.  Obtaining  in  its  simplest  form  the 
expression  for  this  sum  is  often  called  finding  the  sum  of 
the  series. 


170  SECOND  COURSE  IN  ALGEBRA 

The  expression  for  the  sum  is  derived  as  follows : 

S^  =  a-{-ar-\-  ar  -\ h  ar^-^  +  ai^-^  +  ai^-^.  (1) 

(1) .  r,    rS^  =  ar  +  ai^  +  ur^  -\ \-  ar^-^  +  a/^-i  +  ar«.  (2) 

The  terms  ar,  «r^,  etc.,  up  to  ai^~^  in  the  right  member  of 
(1),  occur  in  the  right  member  of  (2).  Hence  if  (2)  be  sub- 
tracted from  (1),  all  these  terms  vanish,  leaving  only  a  and  ar". 

(1)  -  (2),  S^  -  rS^  =  a-  ar». 

Whence  5„  (1  —  ?')  =  a  —  ar", 

and  a-flr"^ 

EXERCISES 

1.  Find  the  sum  of  the  first  ten  terms  of  5,  —  10,  20,  •  •  -o 

Solution :  ^„  =  ;- 

1  —  r 

By  the  conditions,    a  =  5,  r  =  —  2,  and  n  =  10. 
Substituting,  S^^^  =  '^-^(-^Y  =1705.      • 

2.  Find  the  sum  of  1,  5,  25,  •  •  •  to  seven  terms. 

3.  Find  S^  for  the  progression  —  2,  4,  —  8,  •  •  • . 

4.  Find  S^  for  the  progression  50,  10,  2,  ••  • . 

5.  Find  S^  for  the  progression  180,  —  90,  45,  •  •  •. 

6.  Find  S^  for  the  progression  f,  1,  f,  •  •  •. 

7.  Find  .S'^  for  the  progression  c^,  c^,  c\  •  •  • . 

8.  Find  ^S'^.  for  the  progression  3  V2,  6,  6  V2,  •  •  • . 

9.  Find  S^  for  the  progression  81,  —  27  Vs,  27,  •  •  •. 

10.  Find  5„  for  the  progression  3,  15,  75,  •  •  • . 

11.  Find  S^_2  for  the  progression  2x,  4ic*,  Sx',---. 

a  —  tI 

12.  Show  that  for  a  geometrical  progression  S^  = 

13.  What  will  |100  amount  to  in  three  years,  interest  4%, 
compounded  annually  ?  compounded  semiannually  ? 


PROGRESSIONS  171 

14.  A  rubber  ball  falls  from  a  height  of  40  inches  and  on 
each  rebound  rises  40%  of  the  previous  height.  How  far  does 
'it  fall  on  its  sixth  descent  ?  Through  what  distance  has  it 
moved  at  the  end  of  the  sixth  descent  ? 

15.  A  vessel  containing  wine  was  emptied  of  one  third  of  its 
contents  and  then  filled  with  water.  This  was  done  six  times. 
AN'hat  portion  of  the  original  contents  was  then  in  the  vessel  ? 

16.  At  each  stroke  an  air  pump  withdraws  40  cubic  inches 
of  the  contents  of  a  bell  jar  whose  capacity  is  400  cubic  inches. 
After  every  stroke  the  air  remaining  in  the  jar  expands  and 
completely  fills  it.  What  portion  of  the  original  quantity  of  air 
remains  in  the  jar  at  the  end  of  the  tenth  stroke  ? 

91.  Infinite  geometrical  series.  If  the  number  of  terms  of  a' 
geometrical  series  is  unlimited,  it  is  called  an  infinite  geomet- 
rical series. 

In  the  progression  2,  4,  8,  •  •  •  the  ratio  is  positive  and  greater 
than  1,  and  each  term  is  greater  than  the  term  preceding  it. 
Such  a  progression  is  said  to  be  increasing.  Obviously  the  sum 
of  an  unlimited  number  of  terms  of  an  increasing  geometrical 
progression  is  unlimited.  In  other  words,  by  taking  enough 
terms  the  sum  can  be  made  as  large  as  we  please. 

In  the  progression  3,  |,  |,  •  •  •  the  ratio  is  positive  and  less 
than  1,  and  each  term  is  less  than  the  term  preceding  it.  Such 
a  progression  is  said  to  be  decreasing.  Though  the  number  of 
terms  of  such  a  geometrical  progression  be  unlimited,  the  sum 
is  limited ;  that  is,  the  sum  of  as  many  terms  as  we  choose  to 
take  is  always  less  than  some  definite  number.    The  sum  of  the 

first  3  terms  of  the  series  4  +  2-|-l  +  -|4-^H is  7;  of  4  terms 

is  7|- ;  of  5  terms  is  7f ;  of  6  terms  is  7|- ;  of  7  terms  is  7\^. 
Here,  for  any  number  of  terms,  the  sum  is  always  less  than  8. 

The  formula  5„  =  ^LZL^  qn 

may  be  written  ^n  =  JZT;^  -  Y^'  (2) 


172  SECOND  COURSE  IN  ALGEBRA 

For  the  series  3  +  |  +  |  H , 

Now  ay  =  h  (iy  =  h  (^y  =  j\,  («'  =  a-  consequently 
(^y  becomes  very  small  if  n  is  taken  very  great.  Therefore 
3(^)",  the  numerator  of  the  last  fraction  in  (3),  decreases  and 
approaches  zero  as  n  increases  without  limit.  And  hence  as  the 
denominator  of  the  fraction  remains  -^  while  the '  numerator 
approaches  zero,  the  value  of  the  fraction  decreases  and  ap- 
proaches zero  as  n  increases.  Then  if  /S*  denotes  S^,  where  n 
has  increased  without  limit,  we  may  write 

3 
/So,  approaches or  6. 

This  means  that,  though  n  be  very  large,  the  sum  of  the 
series  3  +  |  +  |  H is  always  slightly  less  than  6. 

The  following  is  a  geometrical  illustration  of  the  preceding  series  : 
In  the  adjacent  figure  triangles  ABC,  DEF,  GUI,  etc.,  are  equi- 
lateral.   DEF  is  formed  by  joining  the  middle  points  of  the  sides  of 

ABC,  etc.  Imagine  this  process 

continued  until   an   unlimited 

number  of  triangles  is  so 

formed.    Now  FE  is  ^  of  AB, 

GH  is  1  of  FE,  ML  is   i   of 

GH,  etc.    Therefore,  if  AB  =  1, 

FE  =  i,GH=  1 ,  ML  =  1 ,  etc. 

Hence  the  perimeter  oi  ABC  is 

3;  oiDEF,^;  oiGHI,^;  etc. 

Thus    the    perimeters    of    the 

successive   triangles  form   the 

progression  3,  f  >  f ,  •  •  •,  the  limit 

of  whose  sum  was  found  to  be  6. 

In  the  general  case,  if  r  is  numerically  less  than  1,  the 
numerical  value  of  fraction approaches  zero  as  n  in- 
creases without  limit.    Under  such  conditions  the  formula 


becomes  S^ 
1  —  r 


PROGRESSIONS  173 

This  means  that  for  r  numerically  less  than  1,  aS'^  approaches 
;  but  for  any  definite  value  of  n  it  is  always  numeri- 
cally less  than  this  number. 

Hence  whenever  we  speak  of  the  sum  of  such  a  series  we  mean 
the  limit  which  the  sum  approaches  as  n  increases  indefinitely. 

EXERCISES 

Find  the  number  which  the  sum  of  the  first  n  terms  of  each 
of  the  following  approaches  as  n  increases  without  limit : 

1.  3,1,^,  .... 

Solution :  .S"*  = 


Substituting,  >S'a 


1 


2.  1,  1,  1,  •••.  4.  3,-1,^,....  5a    5a 

I—  K>.    D  a,  —— }  -r—  }   •  • 

3.  2,-1,1,....        5.  2,  V2,  1,....  4     16 

7.  l,x,x%---,(x<l).  ^,11  ,    ^,, 

^  9.  1,  -J  -T'  •••,  (ic  >  1). 

8.  3,  V3,  1,  •••.  '  X    x^  ^  ^ 

10.  .515151.        Hint.    .515151  =  ^^  +  ^/^Viy  + 


11.  .666.  13.  .3939.  15.  .72121. 

12.  .272727.  14.  25.3636.  16.  .3091091. 

17.  A  flywheel  whose  perimeter  is  5  feet  makes  80  revolu- 
tions per  second.  If  it  makes  99%  as  many  revolutions  each 
second  thereafter  as  it  did  the  preceding  second,  how  far  will  a 
point  on  its  rim  have  moved  by  the  time  it  is  about  to  stop  ? 

18.  The  area  of  the  triangle  ABC  (page  172)  is  |  V3;  of  tri- 
angle DEF,  y\  V3  ;  of  triangle  GHI,  ■^%  Vs,  etc.  Find  the  sum 
of  the  areas  of  all  the  triangles  drawn  as  there  supposed. 

19.  The  square  EFHG  is  formed  by  joining  the  middle  points 
of  the  adjacent  sides  of  the  square  A  BCD  on  page  174. 

If  an  unlimited  number  of  squares  is  so  formed,  the  perime- 
ters of  the  squares  will  form  a  geometrical  progression  the  first 


174 


SECOND  COURSE  IN  ALGEBRA 


D 

G 

C 

N 

/\ 

J/ 

H 

^ 

0 

\/ 

X 

three  terms  of  wliieh  may  be  obtained  as  follows  :  Let  AB  =  2', 

then  EB  =  BF  =  1.   EF  can  then  be  found  from  right  triangle 

EBF)  EL  is  ^  of  EF, 

and  EK  =  EL.    Then 

KL  can  be  found  from 

the   right   triangle 

KEL.  The  perimeters 

of  the  first  three 

squares  can  then  be 

found. 

(a)  Show  that  the 
limit  of  the  sum  of 
the  perimeters  of  all 
of  the  squares  is 

16  +  8  V2. 

(b)  Show  that  the        ^  ^  ^ 
limit  of  the  sum  of  the  areas  of  all  the  squares  is  -8  square  units. 

20.  A  loan  of  S  dollars  is  to  be  repaid  in  four  equal  annual 
payments  of  ^  dollars  each.    Find^  if  money  is  worth  r%. 

Solution :  The  sum  due  at  beginning  of  second  year 

=  S(l  +  r)-p.  (1) 

The  sum  due  at  beginning  of  third  year 

=  lS(l-^r)-p-](l  +  r)-p.  (2) 

The  sum  due  at  beginning  of  fourth  year 

=  {[5(l  +  r)-;>](l  +  r)-p}(l  +  r)-p.  (3) 

The  sum  due  at  beginning  of  fifth  year 

=  [{['^(1  +  r)  -p]  (1  +  r)  -p}(l  +  r)-p^  (1  +  r)  -  p.  (4) 
By  the  conditions  of  the  problem,  (4)  =  0,  for  all  the  debt  has  then 
been  paid.    Setting  (4)  equal  to  zero  and  simplifying, 

5(1  +  ry-p(l  +  r)8-p(l  +  ry-p(l  +  r)-p  =  0.         (5) 
Solving  (5)  for;,,  sCl  +  rY 


(1  +  r)8  +  (1  +  r)2  +  (1  +  r)  +  1 
But  the  denominator  in  (6)  is  a  geometrical  series  whose  sum 
by  formula  (B),  page  170,  is  (LhllinA . 


PROGRESSIONS  175 

Substituting  this  last  in  (6),  p  =    ^^    '^  ^^    •  (7) 

In  the  general  case,  if  we  have  n  annual  payments,  the  exponent 

4  in  (7)  would  be  replaced  by  n,  and  then  p  = ^^ ^ . 

^  ^  ^  -^  (1  +  rT  - 1 

21.  A  loan  of  flOOO  is  to  be  repaid  in  three  equal  annual 
payments,  interest  at  5%.    Find  the  payment. 

22.  A  loan  of  |5000  bearing  interest  at  6%  is  to  be  repaid 
in  five  equal  annual  payments.    Find  the  payment. 

Note.  In  the  study  of  geometrical  progressions  we  have  seen  that 
the  sum  of  the  infinite  series  1  +  x^  +  x^  +  x'^  +  •  ■  •  is  a  definite 
number  when  x  has  any  value  less  than  one.  But  it  has  no  finite 
value  when  x  is  equal  to  or  greater  than  one ;  that  is,  we  have  an 
expression  which  we  cannot  use  arithmetically  unless  x  has  a  prop- 
erly chosen  value.  If  we  were  studying  some  problem  which  in- 
volved such  a  series,  it  would  be  a  matter  of  the  most  vital  impor- 
tance to  know  whether  the  values  of  x  under  discussion  were  such 
as  to  make  the  series  meaningless. 

This  question  of  distinguishing  between  expressions  the  sum  of 
whose  terms  approach  a  limit  or  converge,  and  those  which  do  not, 
has  an  interesting  history.  Newton  and  his  followers  in  the  seven- 
teenth century  dealt  with  infinite  series  and  always  assumed  that 
they  converged,  as,  in  fact,  most  of  them  did.  But  as  more  compli- 
cated series  came  into  use  it  became  more  difficult  to  tell  from  in- 
spection whether  they  meant  anything  or  not  for  a  given  value  of 
the  variable. 

It  was  not  until  the  beginning  of  the  nineteenth  century  that 
Gauss,  Abel,  and  Cauchy,  in  Germany,  Norway,  and  France,  re- 
spectively began  to  study  this  subject  effectively,  and  to  devise  far- 
reaching  tests  to  determine  the  values  of  x  for  which  certain  series 
converge  to  a  finite  limit.  It  is  said  that  on  hearing  a  discussion 
by  Cauchy  in  regard  to  series  which  do  not  always  converge,  the 
astronomer  La  Place  became  greatly  alarmed  lest  he  had  made  use 
of  some  such  series  in  his  great  work  on  Celestial  Mechanics.  He 
hurried  home  and  denied  himself  to  all  distractions  until  he  had 
examined  every  series  in  his  book.  To  his  intense  satisfaction  they 
all  converged.  In  fact,  it  has  often  been  observed  that  a  genius  can 
safely  take  chances  in  the  use  of  delicate  processes,  which  seem  very 
foolish  and  unsafe  to  a  man  of  ordinary  ability. 


CHAPTER  XIII 

LIMITS  AND  INFINITY 

92.  Limits.  The  numerical  value  V  of  the  recurring  decimal 
.666  •  •  •  is  a  variable  depending  on  the  number  of  6's  annexed 
on  the  right.  Every  6  thus  repeated  increases  F,  and  the  num- 
ber of  6's  which  may  be  so  repeated  is  unlimited.  Still  V  always 
remains  less  than  f ,  though  constantly  approaching  nearer  and 
nearer  to  that  value.  Here  the  fraction  f  is  called  the  limit  of 
the  variable  V. 

93.  Definition  of  a  limit.  If  a  variable  V  takes  on  successively 
a  series  of  values  that  approach  nearer  and  nearer  to  a  fixed 
number  L  in  such  a  manner  that  the  numerical  value  of  V  —  L 
becomes  and  remains  as  small  as  we  please,  then  V  is  said  to 
approach  the  limit  L. 

This  may  be  written  limit  of  V  =  L. 
The  symbol  =  gives  us  the  equivalent  notation  V  =  L,  which 
is  read  V  approaches  L  as  a  limit. 

94.  Infinity.  If  a  variable  n  takes  on  in  succession  all  the 
values  1,  2,  3,  4,  ■  •  •,  we  can  conceive  of  no  final  value  for  n, 
since  the  system  of  natural  numbers  is  unlimited.  Here  we 
may  say  n  increases  ivithout  limit,  or  n  becomes  infinite. 

95.  Definition  of  the  term  "  infinite."  If  a  variable  n  becomes 
and  remains  greater  than  any  positive  number  k,  however  great, 
we  say  n  increases  without  limit,  or  n  becomes  infinite. 

The  usual  symbol  for  a  variable  which  has  become  infinite 
is  the  sign  oo,  read  infinity. 

Infinity  is  not  a  number  in  the  sense  in  which  2,  V6,  and 
—  7  are  numbers.    It  is  greater  than  any  number.   Eor  present 

176 


LIMITS  AND  INFINITY 


17T 


i 

i  purposes  it  must  be  regarded  as  a  manner  of  speech  rather 
r"  than  as  a  number  that  can  be  added,  subtracted,  multiplied,  or 

divided,  as  ordinary  numbers  are.    In  fact,  we  cannot  operate 

with  the  symbol  go  as  we  can  with  numbers. 

Note.  Some  idea  of  the  reason  why  we  cannot  regard  oo  as  a  nmn- 
Ixn-,  and  operate  with  it  as  we  do  with  ordinary  numbers,  may  be 
seen  if  we  consider  all  even  numbers,  2,  4,  6,  8,  10,  •  •  • .  Evidently 
they  may  be  continued  as  far  as  we  wish,  but  the  number  of  them 
all  cannot  be  expressed  by  any  integer,  for  it  is  greater  than  any 
number ;  that  is,  it  is  infinite.  But  the  number  of  all  integers,  both 
odd  and  even,  was  also  called  infinite,  and  we  symbolize  both  infini- 
ties by  the  same  sign,  oo.  We  have,  then,  two  infinities  which  are 
equal,  or  at  least  they  are  represented  by  the  same  symbol,  but  one 
contains  the  other.  This  is  contrary  to  the  axiom  which  we  always 
assume  for  finite  numbers ;  namely,  that  the  whole  is  greater  than 
any  of  its  parts.  Surely  it  is  not  strange  that  we  cannot  operate 
freely  with  a  symbol  which  violates  this  fundamental  principle. 

We  may  have  a  negative  infinity  as  well  as  a  positive  one. 
In  order  to  indicate  the 
range  of  values  which  both 
X  and  y  may  take  in  graphi- 
cal work,  the  axes  are  often 
marked  as  in  the  adjacent 
figure. 

A  constant  number,  how- 
ever large,  is  never  spoken 
of  as  infinite. 

If  the  variable  nin-  takes 

^  1 

on  in  succession  the  values  1,  2,  3,  4,  •  •  •,  no  final  value  of  - 

.     .     1  "" 

can  be  imagined.    But  as  n  increases  without  limit  -  becomes 

n 

very  small  and  approaches  nearer  and  nearer  to  zero  without 
actually  becoming  zero. 

In  general,  if  a  in  the  fraction  -  is  any  constant  not  zero, 

and  n  a  variable  increasing  without  limit,  -  approaches  zero  as 

a  limit.    Unfortunately,  in  elementary  mathematics  there   is 


1 

+  00 

'- 

^ 

- 

2". 

1 

0 

-2- 

-00 

1 

178  SECOND  COURSE  IN  ALGEBRA 

not  in  general  use  a  symbol  for  a  variable  whose  limit  is  zero, 
though  such  a  symbol  would  be  a  great  convenience. 

The  student  will  frequently  meet  the  statement  —  equals 
zero.  This  statement  is,  of  course,  meaningless  until  it  has 
been  defined,  but  it  may  properly  be  regarded  as  a  way  of 

saying  that  -  approaches  zero  as  a  limit  when  n  is  indefinitely 
increased. 

96.  Interpretation  of  -•    Division  by  zero  is  excluded  from 

mathematics  for  two  reasons :  (a)  It  is  never  necessary,    (b) 
It  would  give  rise  to  endless  ambiguities  and  difficulties. 

Results  of  the  form  -5  where  a  is  a  constant  not  zero,  fre- 
quently arise.  According  to  the  rules  of  computation,  however, 
such  an  expression  has  no  meaning.    Though  it  is  true  that  - 

is  not  a  definite  nmnber,  results  of  this  form  may  sometimes 
admit  of  interpretation. 

As  an  illustration  of  this,  consider  the  following 


EXAMPLE 


Solve  by  determinants  the  system  j  ^ 


o  1  ^-                                     1  +  43 
Solution :  x  =  — —  =  — —  =  - : 


1 

- 

21 

2 

-1 

1 

-2 

h 

-1 

1 

1 

h 

2 

-2z/  =  l,  (1) 

x-7j  =  2.  (2) 


and  y=  ^     ^   =lLZl  =  l^, 

The  graphs  of  (1)  and  (2)  are  parallel  lines.  For  such  lines 
there  is  no  point  of  intersection  and  consequently  the  sys- 
tem has  no  set  of  roots.    Now  as  the  results  for  x  and  y  are 

of  the  form  ->  the  attempt  at  solution  by  determinants  fails. 


LIMITS  AND  INFINITY  179 

Therefore  the  interpretation  of  these  results  is  that  no  set  of 
roots  exists  for  the  system  (1),  (2). 

In  general,  if  for  any  system  of  linear  equations  the  results 

'  obtained  are  of  the  form  -  >  the  system  has  no  set  of  roots ; 

that  is,  the  system  is  inconsistent. 

If  the  student  meets  the  statement  -  =  cc ,  he  should  regard  it  as  a 

loose  use  of  the  statement  that  -  becomes  infinite  as  n  approaches  0. 

n 

97.  Interpretation  of  -  •   The  fraction  -^ becomes  -  when 

0  ic^  —  4  0 

x  =  2.  For  any  value  of  x  other  than  the  critical  value  2,  the 
fraction  equals  a  definite  number.  Usually  we  are  concerned 
with  the  limit  of  such  expressions  as  the  variable  approaches 

X  —  2 
a  critical  value.    The  limit  for  the  fraction  -r is  easily 

x^  —  4  "^ 

found.    We  assign  to  x  successively  the  values  1.9, 1.99, 1.999, 

1.9999,  etc.    The  corresponding  values  of  the  fractions  are  ^§, 

M§5  i^f fj  i§§§§?  ^*c-    Obviously  these  numbers  approach  the 

limit  i. 

We  may  arrive  at  this  result  more  easily  as  follows  :  For  all 

X  —  2 
values  of  x  except  2  the  terms  of  the  fraction  —^ may  be 

divided  by  x  ~  2,  obtaining •     This  result  is  true,  how- 

X  -j—  Zi 

ever  little  x  may  differ  from  2.    Now  if,  without  giving  x  the 

value  2,  we  make  it  approach  2  as  a  limit, will  approach 

^  +  ^  x  —  2 

^  as  a  limit,  and  this  is  the  limit  of  the  original  fraction  -^ — - 

as  well. 

By  either  of  the  preceding  methods  it  can  be  shown  that 
^2  _  g  Q 

?  which  becomes  -  f or  x  =  3,  has  6  as  its  limit.    These 

X  —  6  0 

two  fractions  are  simple  illustrations  of  the  important  fact  that 

the  symbol  -  is  not  a  definite  number.    The  truth  of  this  can 

be  seen  more  clearly  from  a  study  of  the  graph  on  the  fol- 
lowing page. 


180 


SECOND  COURSE  IN  ALGEBRA 


Let 
Then 


y  = 


3 


?/  (x  -  3)  -  (x  +  3)  (ar  -  3)  =  0. 

(3^-x-3)(x-3)  =  0. 

Therefore  x  —  3  =0, 

and  1/  —  X  —  3  =  0. 


(1) 

(3) 
(4) 
(5) 
(6) 


The  graphs  of  (5)  and  (6)  are  given  in  the  following  figure.    To 

understand  what  follows,  it  must  be  remembered  that  y  and  the 

^2  _  9 
fraction are  identical,  and 

X  —  6 

that  the  graphs  (5)  and  (6)  are 
the  complete  graph  of  the  equa- 
tion (1). 

For  every  value  of  x  except  3 
there  is  always  one  value  of  y, 
and  that  value  is  the  ^/-distance 
of  some  point  on  line  (6).  For 
X  =  'S,  however,  the  value  of  y 
is  the  y-distance  of  any  point  on 
line  (5).  Hence  the  fraction 
x2-9 


is  indeterminate  for  x  =  3. 


"^ 

■ 

/tf 

)— 

— 

/ 

/ 

/ 

y 

/ 

(5) 

. 

/ 

-X- 

/ 

v 

/ 

0 

: 

I 

3    ^ 

/ 

'' 

-^ 

■' 

It  is  worth  noting  that  the 

x^  —  d 

limiting  value  of  the  fraction is  seen  from  the  graph  to  be  G, 

X  —  3 

the  y-value  of  the  point  of  intersection  of  lines  (5)  and  (6). 

Note.  The  study  of  the  limiting  value  of  the  ratio  of  two  func- 
tions which  for  certain  values  of  the  variable  takes  on  the  indeter- 
minate form  0/0  was  undertaken  by  the  Frenchman,  L'Hospital,  in 
1696,  and  was  carried  further  by  John  Bernouilli  a  few  years  later. 
A  complete  comprehension  of  the  difficulties  which  surround  this 
subject  has  been  very  slowly  gained  by  mathematical  writers,  and 
even  to-day  it  is  possible  to  find  books  in  which  grave  errors  are 
made  regarding  the  meaning  of  these  expressions. 

The  questions  involved  are  closely  related  to  those  regarding 
the  nature  of  the  infinite  in  mathematics.  The  penetration  of  this 
mystery  is  one  of  the  great  achievements  of  the  latter  half  of  the 
nineteenth  century,  and  to-day  well-informed  mathematicians  have 
as  clear  and  satisfactory  ideas  about  infinite  numbers  as  they  do 
about  ordinary  integers. 


LIMITS  AND  INFINITY 


181 


As  a  final  illustration  that  -  may  have  any  value,  consider 
the  following 


EXAMPLE 


Solve  by  determinants  the  system 


x-y  =  l, 


(1) 


Solution :  x  = 


1  -1 

-3  3 

1  1 

-3  -3 


-Sx-\-37/  =  -S.     (2) 


3-3 


0 


3  +  3 
0 


Now  the  graphs  of  equations  (1)  and  (2)  are  coincident  lines. 
Therefore  any  set  of  values  of  x  and  y  which  satisfies  (1)  will  also 

satisfy  (2),  a  condition  indicated  by  the  indeterminate  result  -  for 
the  unknowns. 


0 


In  general,  if  the  solution  of  a  system  of  linear  equations  in 

two  or  more  unknowns  gives  -  as  values  of  the  unknowns,  the 

system  has  an  infinite  number  of  sets  of  roots;  that  is,  the 
system  is  indeterminate. 

The  symbol  -  then  is  a  symbol  of  indetermination. 


EXERCISES 

Solve  by  determinants  and  interpret  results  : 
x-y  =  l,  x-y=0, 

'  2y-2x=-2.  '  y-x  =  3. 

X  -\-  y  +  z  =  0,  X  -{-y  -\-  z  =  1, 

3.  x-2y-{-3z  =  l,  A.  x-y-2z  =  2, 

2x-y  +  4.z  =  l.  0x  +  0y  +  0z==0. 

5.  From  the  results  obtained  in  Exercise  4  what  conclusion 
is  warranted  regarding  the  number  of  sets  of  roots  belonging  to 
a  system  of  two  linear  equations  in  three  variables  ? 


182  SECOND  COURSE  IN  ALGEBRA 

6.  x  +  i/  +  z  =  2,0x-\-0i/  +  0z  =  0,0x-\-0i/  +  0z  =  0. 

7.  What  do  the  results  obtained  in  Exercise  6  show  in  regard 
to  the  number  of  sets  of  roots  belonging  to  one  equation  in  thnui 
variables  ? 

What  limit  does  each  of  the  following  expressions  approach 
as  n  becomes  oo  ? 


8.i. 

n 

10  J. 

n 

"•»:2- 

-t 

9.?. 

n 

-»:i- 

13.  ''  +  ^ 
n 

n 

15. 


n(n-\-l)  71(72.  +  !)  (^  +  2) 

o  *  AO.  o 


What  limit  does  each  of  the  following  expressions  approach 
as  71  =  0  ? 

14  6  n 

17.  -•  18.  -.  19.  -•  20.  2  7^1  21.  -z- 

nil  w 


Find  the  limit  of : 

^«    1  —  ^  .  H  «„    ic^  —  5  a?  +  6  .   _ 

22. 5  as  X  =  1.  23.  ^ ;; as  a;  =  2. 

1  —  x^.  x^  —  4: 

X  —  2 

24.  -^ 3  as  ic  =  2. 

x^  —  S 


CHAPTER  XIV 

LOGARITHMS 

98.  Introduction.  Logarithms  were  invented  to  shorten  the 
work  of  extended  numerical  computations  which  ^involve  one 
or  more  operations  of  multiplication,  division,  involution,  and 
evolution.  They  have  decreased  the  labor  of  computing  to* 
such  an  extent  that  many  calculations  which  would  require 
hours  without  logarithms  can  be  performed  by  their  aid  in 
one  tenth  of  that  time. 

A  logarithm  is  an  exponent.  A  table  of  common  logarithms 
is  a  table  of  exponents  of  the  number  10.  The  greater  portion 
of  these  exponents  are  approximate  values  of  irrational  num- 
bers. It  follows,  then,  that  computation  by  means  of  loga- 
rithms gives  only  approximate  results.  Tables  exist,  however, 
in  which  each  logarithm  is  given  to  twenty  or  more  decimals ; 
hence  practically  any  desired  degree  of  accuracy  can  be  obtained 
by  using  the  proper  table. 

It  can  be  proved  that  the  laws  given  on  pages  89-90,  govern- 
ing the  use  of  rational  exponents,  hold  for  irrational  exponents. 
In  the  work  on  logarithms  this  fact  will  be  assumed. 

99.  Graphical  explanation  of  logarithms.  The  theory  of  com- 
putation by  logarithms  is  simple,  yet  considerable  time  is 
needed  to  master  its  practical  details.  These  details  and  the 
fact  that  a  logarithm  is  an  exponent  will  be  grasped  more 
readily  if  the  student  gets  from  a  graph  a  first  view  of  the 
meaning  and  use  of  logarithms.  For  this  we  shall  construct 
the  graph  of  the  logarithmic  or  exponential  equation, 

N=  10^. 

183 


184 


SECOND  COURSE  IN  ALGEBRA 


In  this  equation  N  represents  any  positive  number,  and  Z, 
the  exponent  of  10,  is  its  common  logarithm. 

It  will  be  more  convenient  to  assign  values  to  L  and  com- 
pute the  corresponding  values  of  N,  than  to  use  the  reverse 
process.  Moreover,  we  shall  restrict  L  to  values  from  + 1 
to  —  1  inclusive,  and  to  such  fractional  values  that  N  can  be 
obtained  by  the  use  of  square  root. 


First, 
and 
Also 

Similarly, 
Now 
.  Again, 


10^  =  10,  lO''  =  1,  and  10"  ^  =  .1, 
10*  =  VlO  =  3.16227  +. 


10*  =  (10*)*  =  VVIO  =  V3.16227 
10*  =  (10*)*  =  V^-^10  =  -s/iJfs 


1.778  +. 

1.33 +. 

10*  =  (10*)  (10*)  =  (3.16227)  (1.778)  =  d.62  +. 


10*  =  (10*)  (10*)  =  (1.778)  (1.33)  =  2.37  +• 
In  like  manner, 

10^  =  (10*)  (10*)  =  4.21  +. 
Lastly,  10*  =  (10*)  (10*)  =  7.49  +. 
Tabulating  the  values  of  N  and  L  just  obtained,  gives 


L 

0 

i 

.    i 

f 

J 

1 

1 

7 

1 

N 

1 

1.3? 

1.78 

2.37 

3.16 

4.21 

5.62 

7.49 

10 

Since 
the  value  of 


10-^ 
10-* 


10^ 
1 


10* 


10* 

^  =  .133. 


10*      10*  10*       1^ 

Similarly,  10"*  =  -^  =  ~  =  .1778. 

10*       1^ 

In  this   manner  we  obtain  from  the  preceding  table  the 

following  one  for  negative  values  of  L  between  0  and  —  1. 


L 

-i 

-i 

-1 

-i 

_   5 

-1 

_  7 
8 

-1 

N 

.749 

.562 

.421 

.316 

.237 

.178 

.133 

.1 

LOGARITHMS 


185 


— 

"^ 

^ 

^ 

^ 

A^ 

>^ 

^ 

X 

— 

2 

^ 

^ 

< 
-.5g 

/ 

/ 

/ 

r^ 

/ 

•3  i:^ 

/ 

-- 

^T - 

/ 

^ 

y 

L 

AX 

sc 

)Fr 

OUIV 

IBE 

RS 

_  h 

J — 

"7 

1 

— 

' 

> 

3 

^ 

1 

'o^'^ 

/ 

/ 

1 

<• 

-.b- 

' 

' 

° 

»  N 

t 

0 

2 

0 

: 

0 

4 

0 

- 

0 

6 

0 

7 

0 

8 

0 

s 

0 

1 

r 

L 

1 

186  SECOND  COURSE  IN  ALGEBRA 

From  the  values  of  L  and  N  in  the  foregoing  tables  tlic 
logarithmic  (or  exponential)  curve  on  the  preceding  page  is 
constructed.  As  only  positive  values  of  N  are  considered,  the 
curve  never  reaches  the  Z--axis.  The  approximate  values  of 
L  for  numbers  from  .1  to  10  are  measured  from  ON  to  the 
curve. 

From  the  curve,      log  2  =  .3 ;  that  is,  2  =  10«. 

Then  20  =  10  •  2  =  10^  •  lO-^  =  lO^-^. 

Thus  the  logarithm  of  20  is  1  greater  than  the  logarithm  of  2. 
Similarly,  the  logarithm  of  30  is  1  greater  than  the  logarithm 
of  3,  and  so  on. 

Therefore,  if  line  O'N^  be  drawn  one  unit  below  ON,  the 
logarithms  of  numbers  from  10  to  100  are  the  values  of 
distances  to  the  curve  from  points  on  O'iV'  which  correspond 
to  these  numbers.  This  practically  gives  us  a  considerable 
portion  of  the  curve  beyond  point  A. 

EXERCISES 

Find  from  the  curve  the  logarithm  of : 


1.  2. 

6.  7.4. 

11.  .5. 

16.  96. 

2.-3. 

7.  9.6. 

12.  .9. 

17.  100. 

3.  4. 

8.  11. 

13.  25. 

18.  10. 

4.  5. 

9.  15. 

14.  50. 

19.  1. 

5.  6.2. 

10.  .2. 

15.  64. 

20.  32. 

Find  from  the  curve  the  number  whose  logarithm  is  : 

21.  .3.  25.  .95.  29.  -.25.  33.  1.8. 

22.  .4.  26.  -.1.  30.  1.3.  34.  2-. 

23.  .6.  27.  -.4.  31.  1.7.  35.  .84. 

24.  .7.  28.  -.5.  32.  1.6.  36.  1.2. 

The  preceding  exercises  should  familiarize  the  student  with 
the  meaning  of  the  curve.  We  shall  now  use  it  to  explain 
logarithmic  multiplication,  division,  raising  to  a  power,  and 


LOGARITHMS  18T 

extracting  a  root.  It  must  be  remembered  that  the  curve  is  on 
too  small  a  scale  to  give  very  close  approximations.  To  draw 
a  logarithmic  curve  which  would  give  results  sufficiently  accu- 
rate for  most  practical  purposes  would  require  a  piece  of  cross- 
ruled  paper  about  two  feet  square. 

100.  Logarithmic  multiplication.    Multiplication  by  means  of 
the  logarithmic  curve  is  illustrated  in  the 

Example  :  Multiply  3  by  2. 

Solution :  From  the  curve,  log 3  =  .47 ;  hence  3  =  lO*"^. 
From  the  curve,  log  2  =  .3  ;  hence  2  =  10-^. 

Then  3  .  2  =  lO'*' .  10-3  =  lO-'^. 

From  the  curve,  lO-"^^  =  6. 

EXERCISES 
Compute  as  in  the  preceding  example : 

1.  2-5.  4.  .8-8.  7.  .5-80.  10.  .6-80. 

2.  3-3.  5.  30-2.  8.  22- 4.8.  11.  .7-97. 

3.  44.  6.  25-4.  9.  14- 6.  12.  1.5-7.2. 

101.  Logarithmic  division.   Division  by  means  of  the  logarith- 
mic curve  is  illustrated  in  the 

Example :  Divide  40  by  8. 

Solution  :  From  the  curve,  log  40  =  1.6  ;  hence  40  =  IQi-^. 
From  the  curve,  log  8  =  .9  ;  hence  8  =  10-^. 

Then  40  ^  8  =  IQi-^  -  lO-^  =  lO-^. 

From  the  curve,  10-'^  =  5. 

EXERCISES 

Compute  as  in  the  preceding  example : 

1.  8 --2.  5.  16 --.8.  4.6 

2.  6 --3.  6.  48       ^  ^' 


3.  40 --5.  7.  22 

4.  18-^8.  8.  56 -H  8.  *"'      25 


6.  3 

11-  '  ,^    80- 40 

10. 


188  SECOND  COURSE  IN  ALGEBRA 

102.  Logarithmic  involution.    A  number  is  raised  to  a  power 
by  means  of  the  logaritlimie  curve,  as  in  the 

Example :  Find  2\ 

Solution :  From  the  curve,  log  2  =  .3  ;  hence  2  =  10-*. 
Therefore  28  =  (10-8)8  =  10». 

From  the  curve,  10*  =  8. 

1  '^ 

EXERCISES 
Compute  as  in  the  preceding  example : 
1.  3^  4.  41  22.38 


6. 


2.  51  5^.108  ■      6 

3.  3^  ^*      252    *  7.  42. 3^ 

103.  Logarithmic  evolution.    Roots  are  extracted  by  means  of 
the  logarithmic  curve,  as  in  the 

Example:  Find  -v/io. 

Solution  :  From  the  curve,  log  40  =  1.6  ;  hence  40  =  lO^-^. 

Therefore  -v^  =  (40)*  =  (lO^-*)*  =  lO-^a. 

From  the  curve,  lO-'^  =  3.45,  which  is  approximately  Vio. 


EXERCISES 
Compute  as  in  the  preceding  example : 

1.  Vs.  3.  VSI.  5.  7*.  y    38.  V2 

2.  VI.  4.  6*.  6.  52.  V3.  '^ 

From  the  foregoing  work  the  student  should  see  that  a 
logarithm  is  an  exponent,  and  that  by  the  use  of  logarithms 
multiplication  is  effected  by  addition,  division  by  subtraction, 
involution  by  a  single  multiplication,  and  evolution  by  a  single 
division.  The  values  of  N  and  L,  which  up  to  this  time  have 
been  taken  from  the  curve,  will  hereafter  be  obtained  much 
more  accurately  from  the  table  on  pages  200-201. 


LOGARITHMS  189 

104.  Steps  preceding  computation.  Before  computation  by 
means  of  the  table  can  be  taken  up,  two  processes  requiring 
considerable  explanation  and  practice  must  be  mastered. 

I.   To  find  fi'oyn  the  table  the  logarithm  of  a  given  nu7nbe7\ 
II.   To  find  from  the  table  the  number  corresponding  to  a  given 
logarithm. 

105.  Base.  If  iV^  =  b^,  the  logarithm  of  N  to  the  base  b  is  L. 
This  last  is  expressed  by  the  equation  log^N  =  L.  Therefore 
N=b^  and  logj, N=L  are  two  ways  of  expressing  the  same  fact. 

Consequently  2  =  10-^^'^  and  logio2  =  .301  are  equivalent  statements. 

The  base  of  the  common  or  Briggs  system  of  logarithms  is  10. 
The  base  10  is  often  omitted.  Thus  log  2  means  logjQ2.  This 
system  is  used  in  numerical  work  to  the  exclusion  of  all  others. 

The  base  of  the  natural  system  of  logarithms  is  the  irrational 
number  2.7128  -\-,  which  is  usually  denoted  by  e. 

The  natural  system  of  logarithms  is  used  for  theoretical 
purposes  only. 

EXERCISES 

Write  in  the  notation  of  logarithms  : 


1.  300  =  102-^^. 

3. 

4  = 

=  lO-^^ 

5.  .10  =  10-1. 

2.  Q>b  =  10^«i. 

4. 

1  = 

=  10". 

6.  1730  =  10^-28^ 

7.  173  =  102-238. 

9. 

.173  =  10-1 +  -238. 

8.  1.73  =  10-238. 

10. 

.0173  =  10-2+238. 

Write  as  powers  of  10 

11.  log  3  =  .47. 

14. 

log  490  =  2.69. 

12.  log  20  =  1.301. 

15. 

log  .0049  =-3 +  .69. 

13.  log  4.9  =  .69. 

16. 

log  381  =  2.58. 

106.  Characteristic  and  mantissa.  Unless  a  number  is  an 
exact  power  of  10,  its  logarithm  consists  of  an  integer  and  a 
decimal. 

The  integral  part  of  a  logarithm  is  called  its  characteristic. 

The  decimal  part  of  a  logarithm  is  called  its  mantissa. 


190  SECOND  COURSE  IN  ALGEBRA 

The  word  "  mantissa"  means  an  addition ;  that  is,  a  decimal  por- 
tion which  is  added  on  to  the  integral  part,  or  characteristic,  of  the 
logarithm.  This  term  was  used  at  one  time  to  indicate  the  decimal 
part  of  any  number,  but  for  over  a  century  it  has  been  applied  almost 
exclusively  to  logarithms. 

Log  200  =  2.301.  Here  2  is  the  characteristic  and  .301  is 
the  mantissa. 

Biographical  Note.  John  Napier.  Although  many  scientists  have 
been  honored  with  titles  on  account  of  their  discoveries,  very  few  of 
the  titled  aristocracy  have  become  distinguished  for  their  mathematical 
achievements.  A  notable  exception  to  this  rule  is  found  in  John  Napier, 
Lord  of  Merchiston  (1550-1617),  who  devoted  most  of  his  life  to  the  prob- 
lem of  simplifying  arithmetical  operations. 

Napier  was  a  man  of  wide  intellectual  interests  and  great  activity. 
In  connection  with  the  management  of  his  estate  he  applied  himself 
most  seriously  to  the  study  of  agriculture,  and  experimented  with  various 
kinds  of  fertilizer  in  a  somewhat  scientific  manner,  in  order  to  find  the 
most  effective  means  of  reclaiming  soil.  He  spent  several  years  in  theo- 
logical writing.  When  the  danger  of  an  invasion  by  Philip  of  Spain  was 
imminent  he  invented  several  devices  of  war.  Among  these  were  power- 
ful burning  mirrors,  and  a  sort  of  round  musket-proof  chariot,  the  motion 
of  which  was  controlled  by  those  within,  and  from  which  guns  could  be 
discharged  through  little  portholes. 

But  by  far  the  most  serious  activity  of  Napier's  life  was  the  effort  to 
shorten  the  more  tedious  arithmetical  processes.  He  invented  the  first 
approximation  to  a  computing  machine,  and  also  devised  a  set  of  rods, 
often  called  Napier's  bones,  which  were  of  assistance  in  multiplication. 
His  crowning  achievement,  however,  was  the  invention  of  logarithms,  to 
which  he  devoted  fully  twenty  years  of  his  life. 

No  characteristics  are  given  in  the  table  on  pages  200-201. 
The  characteristic  of  any  number  is  obtained  from  an  inspec- 
tion of  the  number  itself  according  to  rules  which  will  now  be 

derived. 

10*  =  10000;  that  is,  the  log  10000  =  4. 

10»  =  1000;  that  is,  the  log  1000  =  3. 

10^  =  100;  that  is,  the  log  100  =  2. 

10^  =  10;  that  is,  the  log  10  =  1. 

10-  ^  =  .1 ;  that  is,  the  log  .1  =  -  1. 

10-  -  =  .01 ;  that  is,  the  log  .01  =  -  2. 

10-  3  =  .001 ;  that  is,  the  log  .001  =  -  3. 


JOHN  NAPIER 


LOGARITHMS  191 

The  preceding  table  indicates  between  what  two  integers 
the  logarithm  of  a  given  number  lies.  This  determines  the 
characteristic. 

Since  542  lies  between  10^  and  10^,  log  542  =  2  plus  a 
decimal. 

And  since  .0045  lies  between  10" «  and  10"  2,  log  .0045  =  -  3 
plus  a  positive  decimal  (or  —  2  plus  a  negative  decimal). 

For  the  determination  of  the  characteristic  of  a  positive 
number  we  have  the  rules 

I.   The  characteristic  of  a  number  greater  than  1  is  one  less 
than  the  number  of  digits  to  the  left  of  the  decimal  point. 

II.  The  characteristic  of  a  number  less  than  1  is  negative  and 
numerically  one  greater  than  the  number  of  zeros  between  the 
decimal  point  and  the  first  significant  figure. 

Accordingly  the  characteristic  of  25  is  1 ;  of  2536  is  3  ;  of  6  is  0 ; 
of  .4  is  -  1 ;  of  .032  is  -  2  ;  of  .00036  is  -  4. 

The  table  on  pages  200-201  gives  the  mantissas  of  numbers 
from  10  to  999.  Before  each  mantissa  a  decimal  point  is  under- 
stood. 

The  numbers  5420,  542,  5.42,  .0542,'  and  .000542  are  spoken 
of  as  composed  of  the  same  significant  digits  in  the  same  order. 
They  differ  only  in  the  position  of  the  decimal  point,  and  con- 
sequently their  logarithms  will  differ  only  in  their  character- 
istics. If  the  base  of  the  system  is  10,  however,  such  numbers 
will  have  the  same  mantissa. 

The  last  two  points  are  easily  illustrated  by  any  two  num- 
bers which  have  the  same  significant  digits  -in  the  same  order. 

Log  5.42  =  .734,  or  5.42  =  lO-'-^l      • 
5.42  .  102  =  542  =  lO--^^^  .  192  =  102.734. 
Therefore  log  542  =  2  •  734. 

The  property  just  explained  does  not  belong  to  a  system  of  loga- 
rithms in  which  the  base  is  any  number  other  than  10.  It  is  a  very 
convenient  property,  as  tables  of  a  given  accuracy  are  far  shorter 
when  the  base  is  10  than  they  would  be  with  any  other  base.  For 
example,  the  table  on  pages  200-201   gives  the  mantissas  of  all 


192  SECOND  COURSE  IN  ALGEBRA 

numbers  from  1  to  999.  But  these  mantissas  are  just  the  same  as 
the  mantissas  of  the  three-figure  decimals  from  .001  to  .999,  or  another 
set  of  a  thousand  numbers.  Were  the  base  any  number  other  than 
10,  the  mantissas  of  the  numbers  from  1  to  999  would  be  different 
from  those  of  the  numbers  from  .001  to  .999.  Four  pages  or  more 
would  then  be  required  to  print  a  table  equivalent  to  the  one  which 
is  here  put  on  two. 

107.  Use  of  the  tables.  To  obtain  the  logarithm  of  a  number 
of  three  or  fewer  significant  figures  from  the  tables,  we  have  the 

E-ULE.    Determine  the  characteristic  by  ins2)ection. 

Find  in  column  N  the  first  two  significant  figures.  In  the 
row  with  these  and  in  the  column  headed  by  the  third  figure 
of  the  given  numJjer  find  the  required  mantissa. 

EXERCISES 
Find  the  logarithm  of : 

1.  271.  4.  65.  7.  2.7.  10.  6. 

2.  344.  5.  650.  8.  2700.  11.  932. 

3.  982.  6.  27.  9.  3.  12.  .932. 

Solution :  The  characteristic  of  .932  is  —  1  and  the  mantissa  is 
.9694.  Hence  log  .932  =  -  1  +  .9694.  This  is  usually  written  in  the 
abbreviated  form  1.9694.  The  mantissa  is  always  kept  positive  in 
order  to  avoid  the  addition  and  subtraction  of  both  positive  and  neg- 
ative decimals,  which  in  ordinary  practice  contain  from  three  to  five 
figures.  Negative  characteristics,  being  integers,  are  comparatively 
easy  to  take  care  of.  (The  student  should  note  that  log  .932  is  really 
negative,  being  - 1  +  .9694,  or  -  .0306.) 

13.  .643.  15.  .00267.  17.  .0101. 

14.  .0532.  16.  .00579.  18.  825000. 

108.  Interpolation.  The  process  of  finding  the  logarithm  of 
a  number  not  found  in  the  table,  from  the  logarithms  of  two 
numbers  which  are  found  there,  or  the  reverse  of  this  process, 
is  called  interpolation. 

If  we  desire  the  logarithm  of  a  number  not  in  the  table,  say 
7635,  we  know  that  it  lies  between  the  logarithms  of  7630 


LOGARITHMS  193 

and  7640,  which  aro  given  in  the  table.  Since  7635  is  halfway 
between  7630  and  7640,  we  assume,  though  it  is  not  strictly 
true,  that  the  required  logarithm,  is  halfway  between  their 
logarithms,  3.8825  and  3.8831.  To  find  log  .7635  we  first  look 
up  log  7630  and  log  7640  and  then  take  half  (or  .5)  their  differ- 
ence (this  difference  may  be  taken  from  the  column  headed  I>) 
and  add  it  to  log  7630:    This  gives 

log  7635  =  3.8825  +  .5  X  .0006  =  3.8828. 

Were  we  finding  log  7638,  we  should  take  .8  of  the  differ- 
ence between  log  7630  and  log  7640  and  add  it  to  log  7630. 

The  preceding  solution  illustrates  the  general 

Rule.  Prefix  the  proper  characteristic  to  the  mantissa  of  the 
first  three  significant  figures. 
.  Then  multiply  the  difference  between  this  mantissa  and  the 
next  greater  mantissa  in  the  table  (called  the  tabular  difference, 
colum^n  Jy  (of  the  table)  hy  the  remaining  figures  of  the  number 
preceded  by  a  decimal  point. 

Add  the  product  to  the  logarithm  of  the  first  three  figures 
and  reject  all  decimals  beyond  the  fourth  place. 

In  this  method  of  interpolation  we  have  assumed  that  the  increase 
in  the  logarithm  is  directly  proportional  to  the  increase  in  the  num- 
ber. As  has  been  said,  this  is  not  strictly  true,  yet  the  results  here 
obtained  are  nearly  always  correct  to  the  fourth  decimal  place. 

EXERCISES 

Find  the  logarithm  of : 

1.  4625.  6.  72.543.  11.  .00386. 

2.  364.7.  7.  10.101.  12.  .0007777. 

3.  42.73.  8.  700.35.  13.  3.1416. 

4.  32.75.  .  9.  505.50.  14.  2.71828. 

5.  546.8.  10.  2.0075.  15.  .023456. 

'  109.  Antilogarithms.    An  antilogarithm  is  the  number  corre- 
sponding to  a  given  logarithm.    Thus  antilog  2  equals  100. 


194  SECOND  COURSE  IN  ALGEBRA 

If  we  desire  the  antilogarithm  of  a  given  logarithm,  say 
4.7308,  we  proceed  as  follows:  The  mantissa  .7308  is  found  in 
the  row  which  has  53  in  column  iST,  and  in  the  column  which 
has  8  at  the  toj).  Hence  the  first  three  significant  figures  of 
the  antilogarithm  are  538.  Since  the  characteristic  is  4,  the 
number  must  have  five  digits  to  the  left  of  the  decimal  point. 

Thus  antilog  4.7308  =  53,800.  Therefore  if  the  mantissa  of 
ft  given  logarithm  is  found  in  the  table,  its  antilogarithni  is 
obtained  by  the 

Rule.  Fimd  the  row  and  the  column  in  which  the  given  man- 
tissa lies. 

In  the  row  found,  take  the  two  figures  which  are  in  column 
N  for  the  first  two  significant  figures  of  the  antilogarithm,  and 
for  the  third  figure  the  number  at  the  top  of  the  column  in  wh  ich 
the  mantissa  stands. 

Place  the  decimal  point  as  indicated  by  the  characteristic. 


EXERCISES 

Find  the  antilogarithm  of : 

1.  3.9309. 

^^6.  8.5740-10. 

10. 

4.6345. 

2.  1.8162. 

HiXT.  8.5740  -  10  =  2.5740. 

11. 

6.9232. 

3.  .6284. 

7.  9.7292-10. 

.    12- 

8.2148. 

4.  1.3541. 

8.  4.8136  - 10. 

-13. 

5.7832 

5.  2.5740. 

9.  0.4533. 

14. 

5.9996. 

If  the  mantissa  of  a  given  logarithm,  as  2.5271,  is  not  in 
the  table,  the  antilogarithm  is  obtained  by  interpolation  as 
follows : 

The  mantissa  5271  lies  just  between 

.5263,  the  mantissa  of  336, 
and  .5276,  the  mantissa  of  337. 

Therefore  the  antilogarithm  of  1.5271  lies  between  33.6  and 
33.7.  Since  the  tabular  difference  is  13  and  the  difference  be- 
tween .5263  and  .5271  is  8,  the  mantissa  .5271  lies  -f^  of  the 


LOGARITHMS  195 

way  from  .5263  to  .5276.   Therefore  the  required  antiiogarithm 
lies  ySj  of  the  way  from  33.6  to  33.7. 
Then  antilog  1.5271  =  33.6  +  j%  x  .1. 

33.6  +  .061  =  33.66. 

Therefore  when  the  mantissa  is  not  found  in  the  table,  we 
have  the 

E/ULE.  Write  the  number  of  three  figures  corresponding  to  the 
lesser  of  the  two  mantissas  between  which  the  given  mantissa 
lies. 

Subtract  the  lesser  mantissa  from  the  given  mantissa  and 
divide  the  remainder  by  the  tabular  difference  to  one  decimal 
place. 

Annex  this  figure  to  the  three  already  found  and  place  the 
decimal  point  where  indicated  by  the  characteristic. 

EXERCISES 

Find  the  antilogarithms  of : 

^  1.  1.5723.             ^  5.  T.2586.  9.  9.2654  - 10. 

2.  2.3921.                6.  7.3472  -  10.  10.  .7829. 

3.  0.6690.                7.  9.8527-10.  11.  7.1050-10. 

4.  2.5728.           .  ^  8.  5.9616  -  8.  12.  6.2308  -  10. 

110.  Multiplication  and  division.  Multiplication  by  logarithms 
depends  on  the 

Theorem.  The  logarithm  of  the  product  of  two  mcm^bers  is 
the  stem  of  the  logarithms  of  the  numbers. 

Proof   Let                      log, N^=l^,  (1) 

and.                                      log,iV^  =  Z,.  (2) 

From  (1),                                N^  =  b\  (3) 

From  (2),                               iV^  =  bh.  (4) 

(3)x(4),                            N^N^  =  bh^h.  (5) 

Therefore  log^,  NJ^,^  =  ^i  -f-  k- 


196  SECOND  COURSE  IN  ALGEBRA 

Division  by  logarithms  depends  on  the 

Theorem.   The  logarithm  of  the  quotient  of  two  numbers  is 
the  logarithm  of  the  dividend  Tninus  the  logarithm  of  the  divisor, 

(1) 

(2) 
(3) 


Proof  Let 

log,N^  =  l„ 

and 

^og,N^  =  l,. 

From  (1), 

2\  =  b'^. 

From  (2), 

N^  =  bh. 

(3)^(4), 

%"-- 

Therefore 

EXERCISES 

Perform  the  indicated  operation  by  logarithms 
1.  18-25. 


Solution : 

Adding, 
Antilog 

log  18  =  1.2553 
log  25  =  1.3979 
log  18  -  25  =  2.6532 
2.6532  =  450. 

2.  37-23. 

^6.  386-27. 

10.  2870-3654. 

3.  28-8. 

7.  432-361. 

11.  286.7-2.341, 

4.  9.8-5. 

8.  589-734. 

^12.  3.412- 2.596. 

5.  42-2.2. 

9.  4326-638. 

13.  432 -.574. 

Solution : 

log  432  =  2.6355  = 
log  .574  =  1.7589  = 

2.6355 
9.7589  - 10 

Adding, 
Antilog 

log432..574  =  2.3944  = 
2.3944  =  247.9. 

12.3944  - 10 

Since  the  mantissa  is  always  positive,  any  number  carried  over  from 
the  tenth's  column  to  the  units  column  is  positive.  This  occurs  in 
the  preceding  solution  where  .6  -f  .7  =  1.3,  giving  +  1  to  be  added 
to  the  sum  of  the  characteristics  +  2  and  —  1,  in  the  units  column. 
Mistakes  in  such  cases  will  be  few  if  the  logarithms  with  negative 
characteristics  be  written  as  in  the  9—10  notation  on  the  right. 


LOGARITHMS  19T 

In  the  preceding  example  and  in  others  which  follow,  two  methods 
are  given  for  writing  the  logarithms  which  have  negative  character- 
istics. This  is  done  to  illustrate  those  cases  in  which  the  second  of 
the  two  ways  is  preferable.  It  should  be  understood  that  in  practice 
one,  but  not  necessarily  both,  of  these  methods  is  to  be  used. 

^14.  385 -.617.  17.  .0876 -.673.  20.  675 -.0236. 

15.  541 -.073.  18.  .07325 .  6.384.         21.  .437  •  .0076. 

-16.  37.6. 00835.         19.  .6381  •  .01897.         22.  891  h- 27. 

Solution :  log  891  =  2.9499 

log27  =  1.4314 

Subtracting,         log  (891  -^  27)  =  1.5185 
Antilog  1.5185  =  33. 

23.  96-12.  \6.  489-27.1.  29.  9876^56.78. 

24.  888 --37.  27.  3460-^4.32.         30.  6432  h- 7.81. 

25.  976-- 321.         28.  4697^281.  31.  3.26 -=- .0482. 

Solution  :  log  3.26  =    0.5132  =  10.5132  -  10 

log  .0482=    2.6830=    8.6830-10 

Subtracting,  log  (3.26  -^  .0482)  =    1.8302  =    1.8302  -    0 
Antilog  1.8302  =  67.64. 

^  32.  2.35  --  .0673.    37.  .07382  -f-  68.72.  \      463.2  •  4.78 

40. 

33.  4.86  -  .721.  256  •  372  -  ^^-^ 

34.  .0635  H- .287.  128  9.63  •  .0872 

35.  .2674  H-  3.86.  347 .  (_  625)  .00635 

36.  7635-8692.    ^^'  346  42.  .078-4.267. 

111.  Involution  and  evolution.  Involution  by  logarithms  de- 
pends on  the     ► 

Theorem.  The  logarithm  of  the  mth  power  of  a  number  is 
7)1  times  the  logarithm  of  the  member. 

Proof    Let  \ogf,N=l.  (1) 

Then  ]^=bK  (2) 

Raising  both  members  of  (2)  to  the  mth  power, 


198  SECOND  COURSE  IN  ALGEBRA 

Therefore  logj,  iV*"  =  ml. 

Evolution  by  means  of  logarithms  depends  on  the 

Theorem.    The  logarithm  of  the  real  inth  root  of  a  numher  is 

the  logarithm  of  the  number  divided  by  m. 

Proof   Let  log,iV^=Z.  (1) 

Then  N=U.  (2) 

Extracting  the  mth  root  of  the  members  of  (2), 

i  i        i. 

{Ny^  =  (by  =  h'^-  (3) 

-       I 
Therefore  log(ir)"*  =  - .  (4) 

EXERCISES 

Compute,  using  logarithms : 

1.  (2.73)^ 

.     Solution :  log  2.73  =  .4362. 

Multiplying  by  3,      log  (2.73)3  =  1.3086. 
Antilog  1.3086  =  20.35. 

2.  (6.32)*.         3.  (34.26)1         4.  (6.715)^      _    5.  (.425)1 

Solution :  log  .425  =  1.6284  =  (9.6284  -  10). 

Multiplying  by  3,      log  (.425)^  =  ^.8852  =  (28.8852  -  30). 
Antilog  5.8852  =  .07676. 

Since  the  mantissa  is  always  positive,  we  have  in  the  preceding 
solution  +  1  (from  3  •  .6)  to  unite  with  —  3  (from  3  •  T).  No  con- 
fusion of  positive  and  negative  parts  need  arise,  if  the  logarithms 
are  written  as  indicated  in  the  parenthesis. 

6.  (.352)*.  7.  (.0672)2.       -8>(.003567)^  9.  v'STe. 

Solution :  log  376  =  2.5752. 

Dividing  by  3,  log  "V^m  =  .8584. 

Antilog  .8584  =  7.218  =  v'm. 

10.  ^^583.  -^11,  ^1235.  12.  v^.000639. 

Solution  :  log  .000639  =  4.8055. 

If  one  divided  4.8055  as  it  stands  by  3,  he  would  be  almost  certain 
to  confuse  the  negative  characteristic  and  the  positive  mantissa.  This 


LOGARITHMS 


199 


and  other  difficulties  may  easily  be  avoided  by  adding  to  the  charac- 
teristic and  subtracting  from  the  resulting  logarithm  any  integral 
multiple  of  the  index  of  the  root  which  will  make  the  characteristic 
positive. 

Thus  log.  000639  =  2.8055  -  G. 

Dividing  by  3,      log -5^.000639  =  .9351  -  2. 
Antilog 


14.  V.0007324. 


2.9351  =  .08612 

15.  -i^.002679. 

16.  (38.4)3. 


-V.000639. 

17.  (4.965)i 

18.  (-  6.387)1 


19. 


20. 


A 


283 . 4.627 


21.  ^209. 


(8.423)^ 


^ 


(23.56)^  •  7.384 
(4.623)^ 


li^.  V87  -  </l6^. 

23.  ^VT^/-w-^ 


Note.  The  following  four-place  table  will  usually  give  results  cor-^ 
rect  to  one  half  of  one  per  cent.  Five-place  tables  give  the  mantissa 
to  five  decimal  places  of  the  numbers  from  1  to  9999,  and,  by  inter- 
polation, the  mantissa  of  numbers  from  1  to  99999.  Five-place  tables 
give  results  correct  to  one  twentieth  of  one  per  cent,  an  accuracy  which 
is  sufficient  for  most  engineering  work. 

Six-place  tables  give  the  mantissa  to  six  decimals  for  the  same 
range  of  numbers  as  a  five-place  table.  The  labor  of  using  a  six-place 
table  is  about  fifty  per  cent  more  than  that  of  using  a  five-place  one. 
For  this  reason  and  for  other  reasons  a  six-place  table  is  of  small 
practical  value. 

Seven-place  tables  give  the  mantissas  of  the  numbers  from  1  to 
99999,  and  by  interpolation  give  the  mantissa  of  numbers  from  1 
to  999999.  Seven-place  tables  are  seldom  needed  in  engineering,  but 
are  of  constant  iise  in  astronomy. 

In  place  of  a  table  of  logarithms  engineers  often  use  an  instru- 
ment called  a  "slide  rule."  This  is  really  a  mechanical  table  of  loga- 
rithms arranged  ingeniously  for  rapid  practical  use.  Results  can  be 
obtained  with  such  an  instrument  far  more  quickly  than  with  an 
ordinary  table  of  logarithms,  and  that  without  recording  or  even 
thinking  of  a  single  logarithm.  A  "  slide  rule  "  ten  inches  long  gives* 
results  correct  to  three  figures.  In  work  requiring  greater  accuracy 
a  larger  and  more  elaborate  instrument  which  gives  a  five-figure 
accuracy  is  used. 


200 


SECOND  COURSE  IN  ALGEBRA 


N 

0 

1 

2 

3    4 

5 

6 

7 

8 

9 

D 

10 

0000 

0043 

0086 

0128  0170 

0212 

0253 

0294 

0334 

0374 

42 

11 

0414 

0453 

0492 

0531  0569 

0607 

0645 

0682 

0719 

0755 

38 

12 

0792 

0828 

0864 

0899  0934 

0969 

1004 

10S8 

1072 

1106 

;^> 

13 

1139 

1173 

12m 

1239  1271 

1303 

1335 

1367 

1399 

1430 

32 

14. 

1461 

1492 

1523 

1553  1584 

1614 

1644 

1673 

1703 

1732 

30 

15 

1761 

1790 

1818 

1847  1875 

1903 

1931 

1959 

1987 

2014 

28 

16 

2041 

2068 

2095 

2122  2148 

2175 

2201 

2227 

2253 

2279 

26 

17 

2304 

2330 

2355 

2380  2405 

2430 

2455 

2480 

2504 

2529 

25 

18 

251)3 

2577 

2601 

2625  2648 

2672 

2695 

2718 

2742 

2765 

24 

19 

2788 

2810 

2833 

2856  2878 

2900 

2923 

2945 

2967 

2989 

22 

20 

3010 

3032 

3054 

3075  3096 

3118 

3139 

3160 

3181 

3201 

21 

21 

3222 

3243 

3263 

3284  3304 

3324 

3345 

3365 

3385 

3404 

20 

22 

3424 

3444 

3464 

3483  3502 

3522 

3541 

3560 

3579 

3598 

19 

23 

3617 

3636 

3655 

3674  3692 

3711 

3729 

3747 

3766 

3784 

18 

24 

3802 

3820 

3838 

3856  •  3874 

3892 

3909 

3927 

3945 

3962 

18 

26 

3979 

3997 

4014 

4031  4048 

4065 

4082 

4099 

4116 

4133 

17 

26 

4150 

4166 

4183 

4200  4216 

4232 

4249 

4265 

4281 

4298 

16 

27 

4314 

4330 

4346 

4362  4378 

4393 

4409 

4425 

4440 

4456 

16 

28 

4472 

4487 

4502 

4518  4533 

4548 

4564 

4579 

4594 

4609 

15 

•29 

4624 

4639 

4654 

4669  4683 

4698 

4713 

4728 

4742 

4757 

15 

30 

4771 

4786 

4800 

4814  4829 

4843 

4857 

4871 

4886 

4900 

14 

31 

4914 

4928 

4942 

4955  496^ 

4983 

4997 

5011 

5024 

5038 

14 

32 

5051 

5065 

5079 

5092  5105 

5119 

5132 

5145 

5159 

5172 

13 

33 

5185 

5198 

5211 

5224  5237 

5250 

5263 

5276 

5289 

5302 

13 

34 

5315 

5328 

5340 

5353  5366 

5378 

5391 

5403 

5416 

5428 

13 

35 

5441 

5453 

5465 

5478  5490 

5502 

5514 

5527 

5539 

5551 

12 

36 

5563 

5575 

5587 

5599  5611 

5623 

5635 

5647 

5658 

5670 

12 

37 

5682 

5694 

5705 

5717  5729 

5740 

5752 

5763 

5775 

5786 

12 

38 

5798 

5809 

5821 

5832  5843 

5855 

5866 

5877 

5888 

5899 

11 

39 

5911 

5922 

5933 

5944  5955 

5966 

5977 

5988 

5999 

6010 

11 

40 

6021 

6031 

6042 

6053  6064 

6075 

6085 

6096 

6107 

6117 

11 

41 

6128 

6138 

6149 

6160  6170 

6180 

6191 

6201 

6212 

6222 

10 

42 

6232 

6243 

6253 

6263  6274 

6284 

6294 

6304 

6314 

6325 

10 

43 

6335 

6345 

6355 

6365  6375 

6385 

6395 

6405 

6415 

6425 

10 

44 

6435 

6444 

6454 

6464  6474 

6484 

6493 

6503 

6513 

6522 

10 

45 

6532 

6542 

6551 

6561  6571 

6580 

6590 

6599 

6609 

6618 

10 

46 

6628 

6637 

6646 

6656  6665 

6675 

6684 

6693 

6702 

6712 

9 

47 

6721 

6730 

6739 

6749  6758 

6767 

6776 

6785 

6794 

6803 

9 

48 

6812 

6821 

6830 

6839  6848 

6857 

6866 

6875 

6884 

6893 

9 

49 

6902 

6911 

6920 

6928  6937 

6946 

6955 

6964 

6972 

6981 

9 

50 

6990 

6998 

7007 

7016  7024 

7033 

7042 

7050 

7059 

7067 

9 

'51 

7076 

7084 

7093 

7101  7110 

7118 

7126 

7135 

7143 

7152 

8 

52 

7160 

7168 

7177 

7185  7193 

7202 

7210 

7218 

7226 

7235 

8 

53 

7243 

7251 

7259 

7267  7?75 

7284 

7292 

7300 

7308 

7316 

8 

54 

7324 

7332 

7340 

7348  7356 

7364 

7372 

7380 

7388 

7396 

8 

LOGARITHMS 


201 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

8 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

8 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

8 

58 

7634 

7642 

7649 

7657 

7664 

7672. 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

6 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

6 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

6 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

6 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

6 

72 

8573 

8579 

8585 

8591 

8597 

8^03 

8609 

8615 

8621 

8627 

6 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

6 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

6 

75 

•8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

6 

76 

8808 

8814 

8820 

8825 

8831, 

8837 

8842 

8848 

8854 

8859 

6 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

6 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

6 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

5 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

5 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

5 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

5 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

5 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

5 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

5 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

5 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

5 

93 

9685 

96^9 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

5 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

5 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

5 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

5 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

4 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

4 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

^9991 

9996 

4 

202  SECOND  COURSE  IN  ALGEBRA 

PROBLEMS  m  MENSURATION 

Solve,  using  logarithms  (obtain  results  to  four  figures) : 

1.  The  circumference  of  a  circle  is  2  ttR.  (tt  =  3.1416, 
R  =  radius.) 

^^Find  the  circumference  of  a  circle  whose  radius  is  42 
inches. 

(b)  Find  the  radius  of  a  circle  whose  circumference  is  6843 
centimeters. 

2.  The  area  of  a  circle  is  ttR^. 

(a)  Find  the  area  of  a  circle  whose  radius  is  3.672  feet. 

(b)  Find  the  radius  of  a  circle  whose  area  is  64.37  feet. 

3.  The  area  of  the  surface  of  a  sphere  is  4  irR^. 

'  (a)  The  radius   of  the  earth  is   3958.79  miles.     Find  its 
surface. 

(b)  Find  the  length  of  the  equator. 

4.  The  volume  of  a  sphere  is  -^-r 

o 

(a)  Find  the  radius  of  a  sphere  whose  volume  is  25  cubic  feet. 
^(1))  Find  the  diameter  of  a  sphere  whose  volume  is  85  cubic 
inches. 

5.  If  the  hypotenuse  and  one  leg  of  a 
right  triangle  are  given,  the  other  leg 
can  always  be  computed  by  logarithms. 

In  the  adjacent  figure  let  a  and  c  be 
given  and  x  required. 

Then  x  =  Vc^  -  a"  =  V(6' +  «)(c  -  a). 

Whence  log  x  =  \  log  (c  +  a)-\-  ^  log  (c  —  a). 

(a)  The  hypotenuse  of  a  right  triangle  is  377  and  one  leg 
is  288.    Find  the  other  leg. 

(b)  The  hypotenuse  of  a  right  triangle  is  1493  and  one  leg 

is  532.    Find  the  other  leg. 

s^     I —  * 

6.  The  area  of  an  equilateral  triangle  whose  side  is  s  is  —  V3. 

(a)  Find  in  square  feet  the  area  of  an  equilateral  triangle 
whose  side  is  11.47  inches. 


LOGARITHMS  203 

(b)  Find  the  side  of  an  equilateral  triangle  whose  area  is 
60  square  centimeters. 

7.  The  area  of  a  triangle  =  -ws(s  —  a')(s  —  b)(s  —  c).    Here 

a,  h,  and  c  are  the  sides  of  the  triangle  and  s  = 

A 

(a)  Find  the  area  of  a  triangle  whose  sides  are  12  inches, 
15  inches,  and  19  inches  respectively. 

(h)  Find  the  area  of  a  triangle  whose  sides  are  557,  840, 
and  1009. 

112.  Exponential  equations.  An  exponential  equation  is  an 
equation  in  which  the  unknown  occurs  as  an  exponent  or  in 
an  exponent. 

Many  exponential  equations  are  readily  solved  by  means 
of  logarithms,  since  log  a^  =  x  log  a.  Thus  let  cv^  =  c.  Then 
X  log  a  =  log  c.    Whence  x  =  log  c  h-  log  a. 

MISCELLANEOUS  EXERCISES 

Solve  for  x : 

I.  8-  =  324. 

Solution  :  log  8-^  =  log  324, 

or  a:  •  log  8  =  log  324. 

Whence  .  =  ISi^  .  ^I^JOo  ^  2.75  +  .' 

log  8         .9031 

2.  3^  =  25.  7.  2^  =  64. 

3.  64^  =  4.  8.  42^+1  =  84. 

4.  16^  =  1024.  81^  2X-1 

9  ■  =  27    ^ 

5.  (-2)"  ==64.  ■  3-^-1 

6.  3  =  (1.04)-.  10.  3^+^  =  5-. 

II.  In  how  many  years  will  one  dollar  double  itself  at  3%, 
interest  compounded  annually  ? 

Solution :  At  the  end  of  one  year  the  amount  of  $1  at  3%  is  $1.03 ; 
at  the  end  of  two  years  it  is  $  (1.03)  (1.03)  or  $(1.03)2 ;  at  the  end  of 
three  years  it  is  $(1.03)^;  and  at  the  end  of  x  years  it  is  $  (1.03)*. 

If  X  is  the  number  of  years  required,  (1.03)^  =  2. 


204  SECOND  COURSE  IN  ALGEBRA 

Taking  the  logarithms  of  both  members  of  the  equation, 

X  log  1.03  =  log  2. 
o  1  •  log  2        .3010      „„  -  , 

12.  In  how  many  years  will  |1  double  itself  at  5%,  interest 
compounded  annually  ? 

13.  In  how  many  years  will  any  sum  of  money  treble  itself 
at  4^,  interest  compounded  annually  ? 

14.  In  how  many  years  will  |265  double  itself  at  3^%, 
interest  compounded  annually? 

15.  In  how  many  years  will  |4000  amount  to  |7360.80  at 
5^,  interest  compounded  annually? 

16.  About  300  years  ago  the  Dutch  paid  |24  for  the  island 
of  Manhattan.  At  4^  compound  interest,  what  would  this 
payment  amount  to  at  the  present  time  ? 

17.  In  how  many  years  will  |12  double  itself  at  4%,  interest 
•  compounded  semiannually? 

18.  Show  that  the  amount  of  P  dollars  in  t  years  at  rcfo, 
interest  compounded  annually,  is  P(l  +  ?•)' ;  compounded  semi- 
annually is  Pf  1  +  -  J  ;  compounded  quarterly  is  ^  ( 1  +  7 )  5 
and  compounded  monthly  is  P(  1  4-  ^ )    • 

19.  Find  the  amount  of  |5000  at  the  end  of  four  years, 
interest  at  8%  compounded  (a)  annually;  (l))  semiannually; 
(c)  quarterly. 

20.  A  man  borrows  $6000  to  build  a  house,  agreeing  to  pay 
|50  monthly  until  the  principal  and  interest  at  6%  is  paid. 
Find  the  number  of  full  payments  required. 

21.  If  each  payment  in  Exercise  20  is  at  once  lent  at  6%, 
compounded  annually,  what  will  they  all  amount  to  by  the 
time  the  final  payment  of  $50  is  made? 


LOGARITHMS  205 

22.  From  Exercises  20  and  21  determine  the  total  interest 
received  by  the  money  lender  up  to  the  time  of  the  last  pay- 
ment.   What  rate  per  cent  on  the  original  |6000  is  this  ? 

Find  the  number  of  digits  in : 

23.  {a)  2^3«.5^;  (h)  3^^;  (c)  2^\ 

24.  Can  the  base  of  a  system  of  logarithms  be  negative  ? 
Explain. 

Find  (without  reference  to  the  table)  the  numerical  values  of : 

25.  log3  9.  29.  log2,9. 

26.  log^S.  30.  log^8  4-logg4. 

27.  log3  2.  31.  log^81-log3^27. 

28.  log,  27.  32.  log^  125  +  log,  25  -  log^  5. 

^-33.  log3a)-log,GV)  +  log^9. 
Simplify : 

34.  log  I  +  log  If.  36.  log  \5  •+  log  ao  _  log  s. 

35.  log  ^V  _  log  ||.  37.  2  log  3  +  3  log  2. 

38.  31og4  +  41og3-21og6. 
Solve  for  x : 

39.  a^  =  (f-\  43.  e"  =  e"*. 

40.  a^-^  =  (f-^.  44.  «^  +  ^  =  Z»2a:  _j_  ^-i_ 

41.  a"-  ^■b'^^  c2^.  ^  45.  a*^  -\-Sa^^=6  a^^. 

42.  3^ .  2^  =  6.  '  ^6.  <x^^  +  a''^  =  6  a*^  —  6  a^^ 


Solve  for  x  and  y : 

2^  =  3^  8-^.5^  =  50, 

*'•  3^-^  =  4^'.  2«^.  322'  =  328. 

2a^-2/  =  5,  3^-6^  =  0, 

*°*  3^ .  9^^  =  9^^  3^+1  -  6^  =  0. 

3;rH-2/  =  9,  0.^-2/^  =  0, 
*^-  2^ .  82y  =  4i«.                      ^'  y  -  a;2  =  0. 


206  SECOND  COURSE  IN  ALGEBRA 

Note.  It  is  not  a  little  remarkable  that  just  at  the  time  when 
Galileo  and  Kepler  were  turning  their  attention  to  the  laborious 
computation  of  the  orbits  of  planets,  Napier  should  be  devising  a 
method  which  simplifies  these  processes.  It  was  said  a  hundred 
years  ago,  before  astronomical  computations  became  so  complex  as 
they  now  are,  that  the  invention  of  logarithms,  by  shortening  the 
labors,  doubled  the  effective  life  of  the  astronomer.  To-day  the 
remark  is  well  inside  the  truth. 

In  the  presentation  of  the  subject  in  modern  textbooks  a  logarithm 
is  defined  as  an  exponent.  But  it  was  not  from  this  point  of  view 
that  they  were  first  considered  by  Napier.  In  fact  it  was  not  tiFl  long 
after  his  time  that  the  theory  of  exponents  was  understood  clearly 
enough  to  admit  of  such  application.  This  relation  was  noticed  by  the 
mathematician  Euler,  about  one  hundred  fifty  years  after  logarithms 
were  invented. 

It  was  by  a  comparison  of  the  terms  of  certain  arithmetical  and 
geometrical  progressions  that  Napier  derived  his  logarithms.  They 
were  not  exactly  like  those  used  commonly  to-day,  for  the  base 
which  Napier  used  was  not  10.  Soon  after  the  publication  (1614)  of 
Napier's  work,  Henry  Briggs,  an  English  professor,  was  so  much 
impressed  with  its  importance  that  he  journeyed  to  Scotland  to  con- 
fer with  Napier  about  the  discovery.  It  is  probable  that  they  both 
saw  the  necessity  of  constructing  a  table  for  the  base  10,  and  to  this 
enormous  task  Briggs  applied  himself.  With  the  exception  of  one 
gap,  which  was  filled  in  by  another  computer,  Briggs's  tables  form 
the  basis  for  all  the  common  logarithms  which  have  appeared  from 
that  day  to  this. 

w,  -^^^  -  "m^^  -^  H''  ^' 


r 


CHAPTER  XV 

RATIO,  PROPORTION,  AND  VARIATION 

113.  Ratio.  The  ratio  of  one  number  a  to  a  second  num- 
ber h  is  the  quotient   obtained  by  dividing  the  first  by  the 

second,  or  -  • 

The  ratio  of  a  to  &  is  also  written  a :  h. 

It  follows  from  the  above  that  all  ratios  of  numbers  are 

fractions  and  all  fractions  may  be  regarded  as  ratios. 

„,       3      c     a  +  h        ,  V2 

1  nus  -  >  — , 5  and are  ratios. 

2    2x    a  —  b  -^ 

The  dividend,  or  numerator,  in  a  ratio  is  called  the  antece- 
dent, and  the  divisor,  or  denominator,  is  called  the  consequent. 

114.  Proportion.  Four  numbers,  a,  b,  c,  and  d,  are  in  propor- 
tion if  the  ratio  of  the  first  pair  equals  the  ratio  of  the  second 

pair. 

a      c 
This  proportion  is  written  a:  b  =  c:  d,  oi-  =  —  ' 

b      d 

Note.  By  the  earlier  mathematicians  ratios  were  not  treated  as  if 
they  were  numbers,  and  the  equality  of  two  ratios  which  we  know 
as  a  proportion  was  not  denoted  by  the  same  symbol  as  other  kinds 
of  equality.  The  usual  sign  of  equality  for  ratios  was  :  :,  a  notation 
which  was  introduced  by  the  Englishman,  Oughtred,  in  1631,  and 
brought  into  common  use  by  John  Wallis  about  1686.  The  sign  = 
was  used  in  this  connection  by  Leibnitz  (1646-1716)  in  Germany, 
and  by  the  continental  writers  generally,  while  the  English  clung  to 
Oughtred's  notation. 

The  first  and  fourth  terms  (ct,  d)  are  called  the  extremes,  and 
the  second  and  third  terms  (b,  c)  are  called  the  means. 

A  mean  proportional  between  two  numbers,  a  and  b,  is  the  num- 
ber m,  if  —  =  —  •    This  means  that  m^  =  ab,  or  m  =  ±  Vo^. 
m       b 

207 


208  SECOND  COURSE  IN  ALGEBRA 

A  third  proportional  to  two  numbers,  a  and  ^,  is  the  number 
...  a      h 

A  fourth  proportional  to  three  numbers,  a,  ^,  and  c,  is  the 
number  /,  if  -  =  -  • 

Since  a  proportion  is  an  equation,  the  axioms,  subject  to 
the  limitations  explained  on  pages  43-44,  apply  to  any  pro- 
portion. 

Then  in  the  proportion  t  =  -^  both  members  may  be  multi- 
plied by  hd^  giving  ad  =  he. 

Therefore,  In  any  ^proportion  the  product  of  the  means  equals 
the  product  of  the  extremes. 

li  ps  =  qr  is  divided  by  qs,  we  obtain 

pi      ^r  p      r 

-e_  =  A- ,     or     —  =  -  •  (1 ) 

q$       ds  q       s 

Also  ps  =  qr  divided  by  rs  gives 


r       s 


And  qr  —  ps  divided  by  pr  gives 

£  =  £. 
p      r 


(2) 


(3) 


Therefore,  If  the  product  of  any  two  numbers  (ps)  equals  the 
product  of  two  other  numhers  (qt),  one  pair  may  he  made  the 
means  and  the  other  pair  the  extremes  of  a  proportion. 

If  -  =  -  )  then  from  (1)  and  (2),  -  =  --•    Here  -  =  -  is  said 
h       d  ^  ''  ^  ^   c      d  c      d 

to  be  obtained  from  -r  =  -  by  alternation. 
b       d    -^ 

If  -  =  -  J  then  from  (1)  and  (3),  -  =  -•    Here  -  =  -  is  said 
b      d  ^^  ^^ac  a       c 

to  be  obtained  from  -  =  -  by  inversion. 
b       d    -^ 


RATIO,  PROPORTION,  AND  VARIATION  209 


^    (        16\   /24      10      1\ 
\         a  /   \a^       a^      a- J 


EXERCISES 

Simplify  the  following  ratios  by  writing  them  as  fractions 
and  reducing  the  fractions  to  lowest  terms 

1.  42:28. 

2.  24a^56al 

3.  ix"-  -  f)  :{x-y).  ^    ( ^      16\  /24  ,  10 

4.  (x^  +  ^y'):(x-\-2y). 

7.  (a)  4  weeks  :  12  hours ;  (b)  480  square  inches  :  2  square 
yards. 

8.  1  mile  :  1  kilometer.    (1  meter  =  39.37  inches^ 

9.  Separate  150  into  three  parts  in  the  ratio  i:6  :  2/ 

10.  If  a  is  a  positive  number,  which  is  the  greater  ratio : , 

,,5  +  3^         5-4^4^^        ,,7- 2a         J-Sa^'^ 
(«)  ir^ — r~  or -^^  ?      (0) z-~  or —  ? 

(c)  If  a  positive  number  is  added  to  or  subtracted  from  both 
terms  of  a  proper  fraction,  what  change  is  produced  in  the 
numerical  value  of  the  fraction  ? 

11  a  :  b  =  c  :  d,  prove  the  following  and  state  the  correspond- 
ing theorem  in  words  : 

11.  a:c  =  b:d.  15.   {a-^b):b=(c  +  d)  :  d. 

12.  b  :  a  =  d  :  c.  16.  (a  +  b)  :  a=  (c  +  d)  :  c. 

13.  «"  :  b'^  =  e»  :  d\  17.   {a  -  b)  :  b  =(e  -  d)  :  d. 

14.  -^^a-.^/b^-VciVd.  IS.  (a-b):a=(G-  d)/b. 

19.  (a  -\-  b)  :  (a  -  b)  =  (G  +  d)  :  (c  -  d).  iX 

The  results  in  Exercises  15, 17,  and  19  are  said  to  be  derived 
from  a:b  =  c  :  d  by  addition,  subtraction,  and  addition  and  sub- 
traction respectively. 

20.  li  a:b  =  c:d=e:f,  prove  that  (a -\- c  +  e)  :  (b -\- d  +/) 
=  a  :b  and  state  the  theorem  in  words. 

21.  Find  a  mean  proportional  between  1.44  and  .0256. 

22.  Find  a  third  proportional  to  15  and  125. 

23.  Find  a  fourth  proportional  to  16^,  8|-,  and  62^. 


210  SECOND  COURSE  IN  ALdEBRA 

24.  Write  5  :  15  =  8  :  24  by  addition,  subtraction,  addition 
and  subtraction,  alternation,  and  inversion. 

Solve : 

25.  8  :  12  =  (3  -  a;)  :  7.  28.  8  :  a;  =  12  :  (10  -  x). 

26.  4  :  a:  =  a;:  169.  29.  3  :  5  =  (a;  -  3)  :  (2  a; -h  18). 
27.3:5  =  ^:2.  30.  20  :  a.  =  .  :  (10  -  .). 

X 

31.  The  surface  of  a  sphere  is  AiTtR^.    Prove  that  for  any 

two  spheres  tt  =  "^  =  7A '  '^'  denoting  the  surface  and  D  the 
diameter. 

4  7r72^ 

32.  The  volume  of  a  sphere  is  — - —    Prove  that  for  any 

•  V       R^      D^ 

two  spheres  -^  =  — ^  =  — ^  >  V  denoting  the  volume  and  D  the 

diameter.         222 

33.  Find  the  ratio  of  the  surfaces  of  the  earth  and  the  moon, 
their  diameters  being  2160  miles  and  7920  miles  respectively. 

34.  Find  the  ratio  of  the  volumes  of  the  earth  and  the  moon. 

35.  y.  ABC  is  any  triangle,  and  KR  is  a  line  parallel  to  BC 
meeting  AB  at  K  and  AC  2it  R,  ^ 

then  the  area  of  ABC  is  to  the 
area  of  AKR  as  AB  :  AK  ,  or  as 

Jc' :  AR\  or  as  BC^ :  Kr\  KA ^^ 

If  in  the  adjacent  figure  the  area    ^ ^^ 

of  ABC  is  100  square  inches,  that 

of  AKR  is  25  square  inches,  and  AB  equals  10  inches,  find  .1  A'. 

36.  If  in  the  figure  of  Exercise  35  ^i?  =  12,  and  triangle 
^  AKR  is  ^  triangle  ABC,  find  AK. 

f  37.  If  in  Exercise  35  the  trapezoid  KRCB  is  eight  times  as 
large  as  triangle  AKR,  and  AC  =  40,  find  AR,"^ 

^  38.  If  AB  equals  32,  and  two  parallels  to  BC  separate  tri- 
angle ABC  into  three  parts  of  equal  area,  find  to  two  decimals 
the  lengths  of  the  three  parts  into  which  AB  i^  divided. 


r 


RATIO,  PKOPORTION,  AND  VARIATION 


211 


39.  If  a  x)lane  be  passed  parallel  to  the  base  of  a  pyramid 
(or  cone)  cutting  it  in  KRL^  then  pyramid  D-ABC  :  pyramid 
D-KRL  =  DH^  :  Ds\  etc. 

If  in  the  adjacent  figure  the 
volumes  of  the  pyramids  are 
4  and  32  cubic  inches  respec- 
tivel}^,  and  the  altitude  DH 
equals  18  inches,  find  DS. 

40.  If  DH  is  12  inches  and 
the  volume  of  one  pyramid  is 
one  half  the  volume  of  the  other, 
find  DS  to  two  decimals. 

V  41.  If  the  volume  of  the  frus- 
tum is  If  of  the  whole  pyramid, 
and  DH  equals  36,  find  DS. 

42.  If  two  planes  parallel  to  the  base  divide  the  whole  pyra- 
'^mid  into  three  parts  having  equal  volumes,  and  DH  =  100,  find 
to  two  decimals  the  parts  into  which  the  planes  divide  DH. 
If  a  :  b  =  e  :  d,  prove  : 


43. 


44. 


45. 


a  —  Sb 

2  ab 


3d 


c  —  3d 

_c^  +  2  d"" 

r  +  d^ 
2cd 


46. 


47. 


48. 


ha" 


5  c^  -  d' 


a^  +  b^ 


+  d' 


3a%-^3  ab'' 
2       ab  +  b'" 


3  c'd  +  3  cd'' 


a 


cd-\-d^ 


3  ab 


3  cd 


49.  In  a  right  triangle  li  is  the  hypotenuse  and  a  and  b  are 
the  legs.  The  corresponding  sides  of  another  right  triangle  are 
H,  A,  and  B.  If  h  :  H  =  a  :A,  prove  a  :  A  =  b  :  B.  Are  the  tri- 
angles similar  ? 

115.  Variatiwi.  The  word  quantity  denotes  anything  which 
is  measurable,  such  as  distance,  rate,  and  area. 

Many  operations  and  problems  in  mathematics  deal  with 
numerical  measures  of  quantities,  some  of  which  are  fixed  and 
others  constantly  changing. 


212  SECOND  COURSE  IN  ALGEBRA 

An  abstract  number,  or  the  numerical  measure  of  a  fixed 
quantity,  is  called  a  constant. 

Thus  the  abstract  numbers  1,  3,  and  —  |  are  constants.  Any 
definite  quantities,  as  the  area  of  a  square  whose  side  is.  2,  the  cir- 
cumference of  a  circle  divided  by  its  diameter  (3.1416  nearly),  the 
time  of  one  revolution  of  the  earth  on  its  axis  (23^,  SG"*,  4.09*),  and 
the  velocity  of  light  through  space  (186,330  miles  per  second)  are 
constants. 

The  numerical  measure  of  a  changing  quantity  is  called  a 
variable. 

For  example,  the  distance  (measured  in  any  unit  of  length)  be- 
tween a  passenger  on  a  moving  car  and  a  point  on  the  track  either 
ahead  of  or  behind  him  is  a  variable,  decreasing  in  the  first  instance, 
increasing  in  the  second.  Other  examples  of  variables  are  one's 
weight,  the  height  of  the  mercury  in  the  thermometer,  and  the  dis- 
tance to  the  sun. 

The  equation  x  =  Si/  may  refer  to  no  physical  quantities 
whatever,  yet  it  is  possible  to  imagine  y  as  taking  oh  in  suc- 
cession every  possible  numerical  value,  and  the  value  of  x  as 
accompanying  every  change,  and  consequently  always  being 
three  times  as  great  as  the  corresponding  value  of  y.  In  this 
sense,  which  is  strictly  mathematical,  x  and  y  are  variables. 

Problems  in  variation  deal  with  at  least  two  variables  so 
related  that  any  change  in  one  is  accompanied  by  a  change  in 
the  other.   Frequently  one  variable  depends  on  several  others. 

For  instance,  the  number  of  lines  of  printing  on  a  page  depends 
on  the  distance  between  the  lines,  the  size  of  the  type,  and  one 
dimension  of  the  page. 

The  symbol  for  variation  is  oc,  and  x  ccy  is  read  x  varies  directly 
as  y,  or  x  varies  as  y. 

116.  Direct  variation.  One  hundred  feet  of  copper  wire  of  a 
certain  size  weighs  32  pounds.  Obviously  a  piece  of  the  same 
kind  200  feet  long  would  weigh  64  pounds ;  a  piece  300  feet 
long  would  weigh  96  pounds,  and  so  on. 

Here  we  have  two  variables  W  (weight)  and  L  (length)  so 
related  that  the  value  of  W  depends  on  the  value  of  L,  and  in 


RATIO,  PROPORTION,  AND  VARIATION  213 

such  a  way  that  W  increases  proportionately  as  L  increases. 
That  is,  W  is  directly  proportional,  or  merely  proportional,  to 
L.  Hence,  if  W^  and  W^  are  any  two  weights  corresponding 
to  the  lengths  L^  and  L^  respectively, 

W^:W,  =  L^:L^.  (1) 

In  the  form  of  a  variation  (1)  becomes 

W  cc  L. 

In  general,  if  xccy,  and  x  and  y  denote  any  two  correspond- 
ing values  of  the  variables,  and  x^  and  y^  a  particular  pair  of 
corresponding  values, 

-  =  -•  (2) 

*i     Si 

From  (2),  ''  =  ©^-  ^^^ 

But  —  is  a  constant,  being  the  quotient  of  two  definite  numbers. 
Call  this  constant  K  and  (3)  may  be  written 

That  is,  if  one  variable  varies  as  a  second,  the  first  equals 
the  second  TnultipUed  by  a  constant. 

Thus  for  the- copper  wire  just  mentioned,  W  =  ^^^  L,  or 
^5  L.  Here,  though  W  varies  as  L  varies,  W  is  always  equal 
to  L  multiplied  by  the  constant  ^^. 

The  phrase  varies  with  is  often  incorrectly  used  in  place  of 
varies  as.  The  latter  should  be  used  to  denote  a  proportional 
change  in  one  variable  with  respect  to  a  second;  the  former 
should  not  be  so  used.  A  boy's  height  varies  with  his  age,  but 
does  not  vary  as  his  age.  At  3  years  the  average  boy  is  about 
3  feet  tall ;  at  12  years  he  is  about  5  feet.  At  the  latter  time, 
if  his  height  varied  as  his  age  from  3  years  up  to  12  years,  he 
would  be  12  feet  tall. 

117.  Inverse  variation.  If  a  tank  full  of  water  is  emptied  in 
24  minutes  through  a  "  smooth "  outlet  in  which  the  area  of 


214  SECOND  COURSE  IN  ALGEBRA 

the  opening,  A,  is  1  square  inch,  an  outlet  in  which  A  is 
2  square  inches  would  empty  the  tank  twice  as  quickly,  or 
in  12  minutes.  And  an  outlet  in  which  ^  is  3  square  inches 
would  empty  the  tank  in  8  minutes. 

Suppose  it  possible  to  increase  or  decrease  A  at  will.  We 
then  have  in  t,  the  time  required  to  empty  the  tank^  and  in  A , 
the  area  of  the  opening,  two  related  variables  such  that  if  A 
increases,  t  will  decrease  proportionally ;  while  if  A  decreases, 
t  will  increase  proportionally.  That  is,  t  and  A  are  inversely 
proportional.  This  means  that  when  A  is  doubled,  t  is  halved ; 
when  A  is  trebled,  t  is  divided  by  3,  and  so  on.  The  relation 
existing  between  the  numerical  values  of  A  and  t  given  in  the 
preceding  paragraph  illustrates  the  truth  of  the  last  statement 
and  of  (1)  which  follows. 

Now  let  t^  and  t,^  be  ani/  two  times  corresponding  to  the 
areas  A^  and  A^  respectively;  then 

t,:t.^  =  A,:A^.  (1) 

The  letters  and  the  subscripts  in  (1)  say :   The  first  time  />• 
to  the  second  time  as  the  second  area  is  to  the  first  area. 
The  proportion  (1)  may  be  put  in  another  form. 
First,  t^-A^  =  t^-A^.  (2) 

Dividing  (2)  by  J,^„       ^  =  ^^^  (3) 

or 


.©-©^ 


:2, 

Whence  ^1-^2  =  ~r''~r'  (-^'^ 

A^    A^ 

Here  the  subscripts  on  the  ^'s  and  those  on  the  ^I's  come  in 
the  same  order. 

In  the  form  of  a  variation  (5)  becomes  t^-7' 

In  general  x  varies  inversely  as  y  when  x  varies  as  the  recip- 
rocal of  y ;  that  is, 

ATxi.  (G) 

y 


RATIO,  PROPORTIO^T,  AND  VARIATION  215 

And  if  X  and  ij  denote  any  two  corresponding  values  of  the 
variable,  and  x^  and  y^  a  particular  pair  of  corresponding  values, 

x:x^  =  -:-'  (7) 

y   Vi 

Whence  —  =  —  ?     or     xy  =  x-^y^.  (8) 

But  x^y^  is  a  constant,  being  the  product  of  two  definite 
numbers.    Call  this  constant  K. 

Then  (8)  becomes  xy  =  K. 

That  is,  if  one  variable  vai'les  inversely  as  another,  the  ])vod- 
uct  of  the  two  is  a  constant. 

118.  Joint  variation.  If  the  base  of  a  triangle  remains  con- 
stant while  the  altitude  varies,  the  area  will  vary  as  the  alti- 
tude. Similarly,  if  the  base  varies  while  the  altitude  remains 
constant,  the  area  will  vary  as  the  base.  If  both  base  and  alti- 
tude vary,  the  area  varies  as  the  product  of  the  two ;  that  is, 
the  area  of  the  triangle  varies  jointly  as  the  base  and  altitude. 
Further,  if  at  any  time  A^  denotes  the  area  of  a  variable  tri- 
angle, and  Aj  and  b^  the  corresponding  altitude  and  base,  and 
if  .4  2  denotes  the  area  at  any  other  time,  and  h^  and  b^  the 
corresponding  altitude  and  base,  then  A^:  A^  =  h^b^ :  hj)^. 

In  the  form  of  a  variation  this  last  becomes 
A  oc  hb. 

In  general,  any  variable  x  varies  jointly  as  two  others,  y  and 

■"'^^  x^yz.  (1) 

If  X  varies  jointly  as  y  and  z,  and  if  x,  y,  and  z  denote  any 
corresponding  values  of  the  variables,  while  x^,  y^,  and  ^^  de- 
note a,  particular  set  of  such  values,  then 

-  =  ^-  (2) 

From  (2),  ''"fe)^^"  ^^^ 


216  SECOND  COURSE  IN  ALGEBRA 

But  in  (3)  the  fraction  — —  is  a  constant,  since  x^,  y^,  and  si^ 

are  particular  values  of  the  variables  x,  y,  and  «.    Calling  this 
constant  K,  we  may  write  x  oc  yz  as  the  equation 

X  =  Kyz, 

One  variable  may  vary  directly  as  one  variable  (or  several  vari- 
ables) and  inversely  as  another  (or  several  others).  Also  one  variable 
may  vary  as  the  square,  or  the  cube,  or  the  square  root,  or  the  recip- 
rocal, or  as  any  algebraic  expression  whatever  involving  the  other 
variable  (or  variables). 

The  theory  of  variation  is  really  involved  in  proportion,  but 
this  fact  is  not  obvious  to  the  beginner.  Hence  it  is  necessary 
to  make  clear  the  meaning  of  the  terms  used  in  variation,  and 
to  show  how  proportion  is  applied  to  the  solution  of  problems 
in  variation.  It  is  doubly  necessary  that  the  student  himself 
make  this  application  in  many  cases,  otherwise  he  will  not 
readily  grasp  numerous  relations  in  physics,  in  chemistry,  and 
in  astronomy ;  for  many  important  laws  of  these  sciences  are 
often  stated  in  the  form  of  a  variation.  In  connection  with 
these  laws  many  problems  arise  which  require  for  their  solu- 
tion clear  notions  of  the  principles  of  variation.  With  a  knowl- 
edge of  proportion  only,  the  student  would  often  find  the  laws 
vague  and  the  problems  difficult. 

{^,/^  PROBLEMS 

1.  If  £c  OC  y,  and  cc  =  4  when  2/  =  6,  find  x  when  y  =  S. 
Solution  :  The  variation  is  direct.    Therefore 

£1  =  ^.  (1) 

Substituting  in  (1),  "^ ""  i-  =  5  *  (2) 

Solving  (2),  x^  =  51. 

2.  If  a;  oc  y,  and  x  =  6  when  y  =  10,  find  y  when  x  —  15. 
!/     Z.  li  xccy,  and  x  =  h  when  y  =  k,  find  y  when  x  =  m. 


1     1 

^1  =  ^2  =  —  =  — 

2/1   2/2 

Substituting  in  (1), 

Q:x,  =  \'.i^. 

Solving  (2), 

x,  =  2. 

RATIO,  PROPORTION,  AND  VARIATION  217 

4.  If  ic  varies  inversely  as  y,  and  a?  =  6  when  y  =  1,  find  x     ^ 
when  ?/  =  21. 

Solution  :  The  variation  is  inverse.    Hence 

(1) 

(2)        • 

x^  =  2. 

5.  If  a?  oc  -  J  and  x  =  4  when  y  =  100,  find  x  when  y  =  10.      ^'  '■  ' 

6.  If  7/  X  -  ?  and  y  =  h  when  «  =  A-,  find  ?/  when  z  =  m. 

7.  If  03  varies  jointly  as  ?/  and  z,  and  a?  =  24  when  y  =  6  and 
,'i;  =  8,  find  X  when  y  =  9  and  ^  =  4. 

Solution:  The  variation  is  joint.    Therefore 

£l     _    2/1^1    _  QN 

•^2         2/2'^2 

Substituting  in  (1),  —  =  — -  •  (2) 

Solving  (2),  x^  =  18.       ^ 

8.  If  X  varies  jointly  as  y  and  z,  and  x  =  3  when  ?/  =  4  and 
=  5,  find  X  when  ^  =  20  and  z  =  2. 

9.  If  £c  varies  directly  as  y  and  inversely  as  z,  and  a:;  =  10 
^vllen  y  =  4  an(i  ;s  =  9,  find  ic  when  y  =  2  and  «  =  6. 

Hint.    Here  a:^ :  Xg  =  —  :  —  • 

^1    ^2 

10.  If  d  varies  directly  as  t^,  and  d  =  64  when  t  =  2,  find  f? 
when  t  =  4:. 

d       t"^ 
Hint.    Here  _1  —  J. . 

d,    q 

11.  If  F  varies  directly  as  T  and  inversely  as  P,  and  F  =  80 
when  P  =  15  and  T=  400,  find  P  when  T  =  450  and  F  =  45. 

12.  The  weight  of  any  object  below  the  surface  of  the  earth 
varies  directly  as  its  distance  from  the  center  of  the  earth.  An 
object  weighs  100  pounds  at  the  surface  of  the  earth.    What 


218  SECOND  COURSE  IN  ALGEBRA 

would  be  its  weight  (a)  1000  miles  below  the  surface  (radius 
of  the  earth  =  4000  miles)  ?  (b)  2000  miles  below  the  surface  '.' 
(c)  at  the  center  of  the  earth  ? 

13.  If  a  wagon  wheel  4  feet  8  inches  in  diameter  makers 
360  revolutions  in  going  a  certain  distance,  how  many  revolu- 
tions will  a  wheel  5  feet  in  diameter  make  in  going  the  same 
distance  ? 

14.  The  distance  which  sound  travels  varies  directly  as  the 
time.  A  man  measures  with  a  stop  watch  the  time  elapsing 
between  the  sight  of  the  smoke  from  a  hunter's  gun  and  the 
sound  of  its  report.  When  the  hunter  was  1  mile  distant,  the 
time  was  4|  seconds.  How  far  off  was  the  hunter  when  the  ob- 
served time  was  2  seconds  ? 

15.  When  the  volume  of  air  in  a  bicycle  pump  is  24  cubic 
inches,  the  pressure  on  the  handle  is  30  pounds.    Later,  when 

-k  the  volume  of  air  is  20  cubic  inches,  the  pressure  is  36  pounds. 
Assume  that  a  proportion  exists  here,  determine  whether  it  is 
direct  or  ^ij^er^^..  and  find  the  volume  of  the  air  when:  the 
pressure  is  48  pounds. 

16.  The  distance  (in  feet)  through  which  a  body  falls  from 
rest  varies  as  the  square  of  the  time  in  seconds.  If  a  body  falls 
16  feet  in  1  second,  how  far  will  it  fall  in  6  seconds  ? 

17.  The  intensity  (brightness)  of  light  varies  inversely  as 
the  square  of  the  distance  from  the  source  of  the  light.  A 
reader  holds  his  book  4  feet  from  a  lamp,  and  later  6  feet 
distant.  At  which  distance  does  the  page  appear  brighter  ? 
how  many  times  as  bright? 

18.  A  lamp  shines  on  the  page  of  a  book  9  feet  distant. 
Where  must  the  book  be  held  so  that  the  page  will  receive 
four  times  as  much  light  ?  twice  as  much  light  ? 

19.  The  weight  of  an  object  above  the  surface  of  the  eartli 
^varies  inversely  as  the  square  of  its  distance  from  the  center 

of  the  earth.    An  object  weighs  100  pounds  at  the  surface  of 


RATIO,  PROPORTION,  AND  VARIATION  219 

the  earth.    What  would  it  weigh  (a)  1000  miles  above  the 
surface  ?  (h)  2000  miles  above  the  surface  ?  (c)  4000  miles  ? 

20.  The  area  of  a  circle  varies  as  the  square  of  its  radius. 
The  area  of  a  certain  circle  is  154  square  inches  and  its  radius 
is  7  inches.  Find  the  radius  of  a  circle  whose  area  is  594  square 
inches. 

21.  The  weight  of  a  sphere  of  given  material  varies  directly 
as  the  cube  of  its  radius.  Two  spheres  of  the  same  material 
have  radii  2  inches  and  6  inches  respectively.  The  first  weighs 
6  pounds,    rind  the  weight  of  the  second. 

22.  The  time  required  by  a  pendulum  to  make  one  vibration 
varies  directly  as  the  square  root  of  its  length.  If  a  pendulum 
100  centimeters  long  vibrates  once  in  1  second,  find  the  time 
of  one  vibration  of  a  pendulum  64  centimeters  long. 

23.  Find  the  length  of  a  pendulum  which  vibrates  once  in 
2  seconds ;  once  in  5  seconds. 

24.  The  pressure  of  wind  on  a  plane  surface  varies  jointly 
as  the  area  of  the  surface  and  the  square  of  the  wind's  velocity. 
The  pressure  on  1  square  foot  is  .9  pounds  when  the  rate  of 
the  wind  is  15  miles  per  hour.  Find  the  velocity  of  the  wind 
when  the  pressure  on  1  square  yard  is  18  pounds. 

25.  The  pressure  of  water  on  the  bottom  of  a  containing 
vessel  varies  jointly  with  the  area  of  the  bottom  and  the  depth 
of  the  water.  When  the  water  is  1  foot  deep,  the  pressure 
on  1  square  foot  of  the  bottom  is  62.5  pounds.  («)  Find  the 
pressure  on  the  bottom  of  a  tank  12  feet  long  and  8  feet  wide 
in  which  the  water  is  6  feet  deep.  (Z').Find  the  total  pressure 
on  one  end  and  on  one  side. 

26.  The  cost  of  ties  for  a  railroad  varies  directly  as  the 
length  of  the  road  and  inversely  as  the  distance  between  the 
ties.  The  cost  of  ties  for  a  certain  piece  of  road,  the  ties  being 
2  feet  apart,  was  $1320.  Find  the  cost  of  ties  for  a  piece  twenty 
times  as  long  as  the  first,  if  the  ties  are  2^  feet  apart. 


220  SECOND  COURSE  IN  ALGEBRA 

27.  The  force  of  gravitation  between  the  earth  and  the  sun 
is  36  X  10"  tons.  Imagine  this  force  replaced  by  a  gigantic 
cable  the  ends  of  which  are  tied  one  to  the  earth  and  the  other 
to  the  sun.  Then  compute  in  miles  the  diameter  of  the  cross 
section  of  the  cable  which  would  just  stand  the  strain,  knowing 
that  a  cable  whose  cross  section  is  1  square  inch  will  support 
60,000  pounds  without  breaking. 

28.  It  has  been  shown  that  if  one  variable  varies  as  another, 
the  second  multiplied  by  a  constant  number  equals  the  first. 
It  is  often  desirable  to  determine  this  constant.  Suppose  such 
to  be  the  case  in  Problem  16. 

Solution:  Since  d:cf,  (1) 

then  d  =  Kt^(K  being  some  constant).         (2) 

But  when  t  =  2  and  d  =  64,  substituting  in  (2)  gives 

64  =  K(2)^  =  4:K;  whence  K  =  16. 

29.  In  Problem  21  IF  oc  r^;  hence  W  =  K?^.    Find  K. 

30.  In  Problem  22  ^  oc  V7.  Find  the  constant  which  multi- 
plied by  V7  gives  t. 

31.  In  Problem  14  find  the  constant  connecting  d  and  t  in 
the  equation  d  =  Kt  and  determine  its  practical  meaning. 

32.  It  has  been  shown  that  if  one  variable  varies  inversely 
as  a  second,  the  product  of  the  two  is  a  constant.  Find  this 
constant  in  Problem  15. 

33.  The  area  of  a  triangle  varies  jointly  as  its  base  and  alti- 
tude.   What  is  the  constant  involved  ? 

34.  The  area  of  a  circle  varies  as  the  square  of  the  diameter. 
What  is  the  constant  involved  ? 

^  35.  The  volume  of  a  sphere  varies  as  the  cube  of  the  diam- 
eter.   What  is  the  constant  involved  ? 

,  36.  Give  concrete  illustrations  of  direct,  inverse,  and  joint 
variation  different  from  those  given  in  this  book. 

37.  lfx^-\-^azx^  —  lf,  prove  x  -\-  y  (xx  —  y. 

38.  If  ic^  +  7/^  oc  ic^  —  ]/,  prove  x  -\-  y  ocx  —  y. 


RATIO,  PROPORTION,  AND  VARIATION  221 

39.  li  X  -\-  y  c/Lx  —  y,  prove  x^  —  xy  -{-  y'^  cj: x^  -i-  xy  -\-  y^. 

40.  In  the  equation  x  =  Ky  assign  some  numerical  value  to 
K  and  graph  the  resulting  equation. 

Direct  variation  between  two  variables  is  represented  by  what 
sort  of  a  curve  ? 

41.  Assign  some  numerical  value  to  iiC  in  cc  =  —  and  then 
graph  the  resulting  equation. 

Inverse  variation  between  two  variables  is  represented  by 
what  sort  of  a  curve  ? 

^   42.  Construct  a  graph  showing  the  relation  between  two 
variables  one  of  which  varies  as  the  square  of  the  other. 

43.  Construct  a  graph  showing  the  relation  between  two 
variables  if  one  varies  inversely  as  the  square  of  the  other. 

44.  The  cost  of  a  certain  kind  of  nails  is  5  cents  x^er  pound. 
If  G  represents  the  cost  of  n  pounds,  what  is  the  equation  which 
connects  c  and  n  ?    What  is  the  graph  of  this  equation  ? 

45.  If  a  quantity  of  gas  is  under  a  changing  pressure,  the 
volume  decreases  as  the  pressure  increases,  and  vice  versa.  Let 
P  denote  the  pressure  and  V  the  volume ;  then  state  an  equa- 
tion expressing  the  relation  between  P  and  V  and  determine 
the  kind  of  a  curve  which  will  represent  this  relation. 

May  negative  pressui-es  or  negative  volumes  be  considered 
here  ? 


CHAPTER  XVI 


IMAGINARIES 


119.  Definitions.  When  the  square  root  of  a  negative  num- 
ber arose  in  our  previous  work,  it  was  called  an  imaginary, 
and  no  attempt  was  then  made  to  use  it  or  to  explain  its 
meaning.  The  treatment  of  imaginaries  was  deferred  because 
there  were  so  many  topics  of  more  importance  to  the  beginner. 
It  must  riot  be  supposed,  however,  that  imaginaries  are  not  of 
great  value  in  mathematics.  They  are  also  of  much  use  in  cer- 
tain branches  of  applied  science;  and  it  is  unfortunate  that 
symbols  which  can  be  used  in  numerical  computations  to 
obtain  practical  results  should  ever  have  been  called  imagi- 
nary. By  such  a  name  something  unreal  and  fanciful  is  sug- 
gested. To  obviate  this  it  has  been  proposed  to  call  imaginary 
numbers  orthotomie  numbers,  but  this  name  has  been  little  used. 

The  equation  cc^  -f- 1  =  0,  or  ic^  =  —  1,  asks  the  question, 
"  What  is  the  number  whose  square  is  —  1  ?  "  By  defining  a 
new  number,  V—  1,  as  a  number  whose  square  is  —  1,  we 
obtain  one  root  for  the  equation  ic^  +  1  =  0.  Similarly,  V—  5 
is  a  number  whose  square  is  —  5.  And,  in  general,  V—  n  is  a 
number  whose  square  is  —  n.  Obviously  V—  5  means  some- 
thing very  different  from  Vs. 

The  positive  numbers  are  all  multiples  of  the  unit  -f- 1,  and 
the  negative  numbers  are  all  multiples  of  the  unit  —  1.  Simi- 
larly, pure  imaginary  nmnbers  are  real  multiples  of  the  imagi- 
nary unit  V—  1. 

Thus  V^  +  V^  =  2  V^,and  V^  +  2  V^  =  3  V^,etc. 
Further,  V^  =  2  V^;  V-a^^aV^;  V^  =  Vs  V^. 


The  imaginary  unit 
that  is,  3  V—  1  =  3  l. 


1  is  often  denoted  by  the  letter 


222 


IMAGINARIES 


223 


If  a  real  number  be  united  to  a  pure  imaginary  by  a  plus 
sign  or  a  minus  sign,  the  expression  is  called  a  complex  number. 

Thus  —  2  +  V—  1  and  3  —  2V—  4  are  complex  numbers.  The  gen- 
eral form  of  a  complex  number  is  a  +  hi,  in  which  a  and  h  may  be 
any  real  numbers. 

Note,  Up  to  the  time  of  Gauss  (1777-1855)  complex  numbers 
were  not  clearly  understood,  and  were  usually  thought  of  as  absurd. 
The  situation  reminds  one  of  the  time  when  negative  numbers  were 
similarly  regarded,  and  the  veil  was  removed  from  both  in  about  the 
same  way.  It  was  found  that  negative  numbers  really  had  a  signifi- 
cance ;  that  they  could  be  used  in  problems  that  involve  debt,  opposite 
directions,  and  many  other  everyday  relations.  The  interpretation 
of  imaginary  numbers  is  not  quite  so  obvious,  but  none  the  less  actual 
and  simple.  As  soon  as  it  was  seen  that  they  could  be  represented 
with  real  numbers  as  points  on  a  plane  (see  page  231)  the  ice  was 
broken,  and  it  needed  only  the  insight  and  authority  of  a  man  like 
Gauss  to  give  them  their  proper  place  in  mathematics. 

120.  Addition  and  subtraction  of  imaginaries.  The  fundamental 
operations  of  addition  and  subtraction  are  performed  on  imagi- 
nary and  complex  numbers  as  they  are  performed  on  real  num- 
bers and  ordinary  radicals  of  the  same  form. 

Thus  2V^+4V^-6V£l, 

and  5V^-3V^  =  2V-1. 

Also         3-^5  V^  -h  4  -  2  V^  =  7  +  3  V^. 
Similarly,  a  ■\-  hi  +  c  ■\-  di  =  a  +  c  -{■  {h  +  d)  i. 


EXERCISES 


1.  3V-l-f4V-l- 

-2V-1. 

6.  5V-36^'-^-2V-49x'^ 

2.  V-4  +  V-9. 

7.  V-  18  -f  V-  8. 

3.  V-25-V-16. 

8.  (-12)^ +  (-27)*. 

4.  5V-l-t-V-9. 

9.  3_|_2V-l+5-6V-l. 

5.  V-4+V-16. 

10.  o^-x'-7a-S^-x\ 

11.  4-8V^ 

i;-f  16- 

-3V-4. 

12.  6  -  2  V-  64a;^  -  3  V-  25cc2  +  8 


224 


SECOND  COURSE  IN  ALGEBRA 


13.  18  -  3(-  1)*  +  6(-  2)*  +  (-  100)^  4-  4. 

14.  5  V^  +  3  V^  -  V-27  +  2  V^. 

15.  6  V-  4  a*  -  7  a^  V^  +  3  V^  -  5  V-  24. 

16.  (12  -  6  V^)  -  (15  +  2  V-  36). 

17.  ?,a-2x-(2a-\/-u}  -hTa^-sT^.   ■ 

18.  {x  —  iy)  —  (n  —  iv). 

Write  as  a  multiple  of  V—  1 : 

19.  V-10.  21.  2V 

20.  V^^. 


23.  aV^. 


24.  V-  a-b. 

121 .  Multiplication  of  imaginaries.   By  the  definition  of  square 
root,  the  square  of  V—  n  is  —  n. 
Therefore 

( viTi)' = ( v:ny  v^ = - 1  v^. 

(V3i)'  =  (v^;(V:n)^  =  (_!)(. 

To  multiply  V—  2  by  V—  3  we  write  V—  2 

and  V^  as  Vs  •  V^. 

Then  V^.  V^  =  (  V2  •  V^)(  VS- V^) 

=  V6- V^.  V^  =  -V6. 
Similarly, 

2  V^(-  3  V^)  =  2  V5 .  V^ (-  3  V2  .  V^)  =  6  VlO. 

In  general,  if  the  V—  a  and  the  V—  b  are  two  imaginaries 
whose  product  ^or  quotient)  is  desired,  they  should  first  be 
written  in  the  form  Va  •  V—  1  and  V^  •  V—  1,  and  the  multi- 
plication (or  division)  should  then  be  performed.  This  method 
will  prevent  many  errors. 

In  this  connection  it  must  be  clearly  understood  that  one 
rule  followed  in  multiplication  of  radicals  (see  page  98)  does 
not  apply  to  imaginary  niunbers. 

Thus  V2  .  V3  =  V2T3  =  V6. 

But  V—  2  •  V—  3  does  not  equal  V(—  2)  (—3),  which  equals  Vs. 


IMAGINARIES 


225 


In  multiplying  two  complex  numbers,  write  each  expression 
in  the  form  a  -\-hl  and  proceed  as  in  the  following 


Multiply  2 
Solution : 

Multiplying, 
Rewriting, 


EXAMPLE 

3  by  3  -  V^. 

2+V^  =  2+V3- V 
3-V^  =  3-V7.  V 


6  +  3  VsV- 
6  +  3  V^ 


1-2  V7V^  +  V21 
■  2  V^  +  V2I 


EXERCISES 

Perform  the  indicated  multiplications  and  simplify  results  : 
(V^)^  12.    ■\/a-\-b-  V-  a-b. 

^  13.  (2  +  V^)(2  -  vCII). 
/  14.  (3  +  V^)(3  -  V^). 
/  15.  (4-2V3^)(4  +  2V3  0. 
16.  (3  + V^)(6-V^). 


8.  V-25 .  Vs. 

9.  2V^.3 


10. 
11. 


V—  m 
4V^(-3 

23.  (x-kjf 

24.  {-\-\-h 


17. 

(4 

-2i)(3-2V3i). 

18. 

(a 

+  ib)  (g  +  id). 

19. 

(a 

+  i^)(a 

+  ib). 

20. 

(a 

+  bi)  (a 

-  bi). 

21. 

-h  +  i- 

/^If. 

3)- 

22. 

1  1  -« 

2  2 

/^f. 

(. 

''  +  iyf. 

- 

sy 

-(- 

■h- 

-iV3 

3)'. 

25.  (a  +  i  Vl  -  x'){a  -  i  Vl  -  £c^). 

Note.  Long  before  the  time  of  Gauss  mathematicians  had  performed 
the  operations  of  multiplication  and  division  on  complex  numbers  by 
the  same  rules  that  they  used  for  real  numbers.  As  early  as  1545 
Cardan  stated  that  the  product  of  5  +  V—  15  and  5  —  V—  15  was  40. 
However,  he  was  not  always  equally  fortunate  in  obtaining  correct 

results,  for  in  another  place  he  sets  -  (  —  a. I  = =  -  • 

4^      N     4;       V64      8 


226 


SECOND  COURSE  IN  ALGEBRA 


Even  the  r3,ther  complicated  formula  for  extracting  any  root  of  a 
complex  number  was  discovered  in  the  early  part  of  the  eighteenth 
century.  But  all  of  these  operations  were  purely  formal,  and  seemed 
to  most  mathematicians  a  mere  juggling  with  symbols  until  Gauss 
showed  clearly  the  place  and  usefulness  of  such  numbers. 

122.  Division  of  imaginaries.  One  complex  number  is  the 
conjugate  of  another  if  their  product  is  real.  Thus  a  +  hi  and 
a  —  bi  are  conjugates.  Conjugate  complex  numbers  are  used 
in  division  of  imaginary  expressions  as  conjugate  radicals  are 
used  in  division  of  radicals. 

Division  by  an  imaginary  is  performed  by  writing  the  dividend 
over  the  divisor  as  a  fraction  and  then  multiplying  both  numer- 
ator and  denominator  by  the  simplest  imaginary  expression 
which  will  make  the  resulting  denominator  real  and  rational. 

EXAMPLES 


1.   V-6- 

^v^. 

Solution : 

V-8-*-V-2  =  - 

/_6     V-e-A 

/-2      V-2.A 
v^-  1      -  Vl2 

/-2 
/-2 

V8. 

V2.  V^- V2- 
2.  3  -^  (2  +  V^). 
Solution  :  3  ^  (2  +  V^)  = 


3  (2  -  V^) 


4  +  3 


(2+V^)(2-V-3) 
6-3V^ 


EXERCISES 
Perform  the  indicated  division  : 


2.  V-  6  --  V^. 

5.  Ve^V^. 


6.  (_  25)*  --  (-  81)*. 

7.  ~\/ax  -f-  V—  X. 

10.  [(_a^)*_(_a^)i]^(. 


ay 


f 


I 


IMAGINARIES 


227 


11.  2-(l- V^). 

12.  3--(2- V^). 

y  13.  2V^-(V:ri  +  3). 

<  14.  3V^-(2V^  +  2). 
15.-1  +  ^^ 


16. 


17. 


18. 


1  +  i 

a 

a  4-  bi 

a  -\-  ib 
c  -\-  id 


_  1  _  V-  3 

19.  (2+-3^)--(2i-l)(5^-3)=? 

20.  Is  ^  V3  -  1  a  cube  root  of  8  ? 

21.  Is  1  -  V^  a  cube  root  of  -  8  ? 

22.  Does  x'-A.x  +  l  =  0,\ix  =  2  ±  V^ ? 

23.  Does  x  =  %  V— 10,  2/  =  —  f  V-  10  satisfy  the  system 
x^'-xy-  12  y^  =  8,  x^  +  xy  -  10  ?/'  =  20  ? 

24.  Determine    whether    the    sum    and    the    product    of 
2  +  3  V—  1  and  2  —  3  V—  1  are  real  numbers. 

25.  Show  that  the  sum  and  the  product  of  any  two  conjugate 
complex  numbers  is  real. 

26.  Show  that  the  quotient  of  two  conjugate  complex  num- 
bers is  complex. 

27.  Point  out  the  error  in  the  followinsr : 
The  equation  Vic  —  y 


V?/  —  X  is  an  identity. 
Let  X  =  a  and  y  =  h,  and  (1)  becomes 

V«.  —  h  =  i  -\/b  —  a. 
Now  let  X  =  b  and  y  =  a,  and  (1)  becomes 


V^  —  a  =  i  -y/a  —  b. 


From  (2)  and  (3), 


■y/a-b-  Vb 


Whence 


.p(^^l,^a--\^a-b). 
i\  or  1  =  -  1. 


(1) 

(2) 

(3) 
(4) 


123.  Equations  with  imaginary  roots.  The  student  should 
now  be  able  to  solve  and  check  equations  which  have  imagi- 
nary roots. 


6a; +  36  =  0. 


3. 

x" 

+  ^x  +  l  = 

0. 

4. 

x" 

-3a;  +  10  = 

=  0. 

5. 

2i 

c2_,_6^_^5 

=  0. 

9. 

a;«  =  27. 

11. 

10. 

a;»  =  -  8. 

12. 

228  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Solve  and  check  the  equations  which  follow : 

1.  a;2  4-  4ic  +  12  =  0.  6.  3a;2  -  7a;  +  6  =  0. 

7.  a;»  =  l. 

Hint.  If  ar' =  1,  ar^  -  1  =  0. 
Hence  (x  —  1)  (a:^  +  x  +  1)  =  0. 
Then  x  -  1  =  0, 

and  a;2  +  a;  +  1  =  0,  etc. 

8;^a;»=8. 

x'  =  l.  13.  a;«  =  l. 

a;*  =  16.  14.  a;«  =  64. 

15.  How  many  square  roots  has  any  real  number  ?  cube 
roots  ?  fourth  roots  ?  sixth  roots  ? 

16.  What  do  Exercises  7-14  indicate  regarding  the  number 
of  Tith  roots  which  any  real  number  may  have  ? 

17.  8a;«  -  27  =  0.  7^  22.  4a;*  -}-  20a;2  +  21  =  0. 
K  18.  125a;^  +  64  =  0.  ^  23.  64a;*  -  12a;2  -  27  =  0. 
f  19.  (a;?  +  5) (x^ -7)+ 27=  0.      24.  9  a;*  +  18  a;^  +  8  =  0. 

20.  a;»  -  a;2  4-  2a;  -  2  =  0.        25.  50a;*  +  135a;2  +  36  =  0. 

21.  a;«  +  7a;^-8  =  0.  26.  (a;^  +  9)  (a;^  +  2  a;  +  8)  =  0- 
f  27.  (a;2  +  xf  +  13(a;2  +  a;)  +  36  =  0. 

28.  (x"  +  5a;)2  +  17 (a;^  +  5a;)+  66  =  0. 

29.  Solve  X  -\-  y  =  4:,  x^  —  ^xy  —  if  =  —  ^^  and  check. 

-^  30.  Solve  z^  +  x'^  =  130,  z  +  x  +  2  ^z  +  x  =  2  and  check. 
31.  Solve  Exercise  4,  page  146,  and  check. 

124.  Factors  involving  imaginaries.  After  studying  radicals 
we  enlarged  our  previous  notion  of  a  factor,  and,  with  certain 
limitations,  employed  radicals  among  the  terms  of  a  factor. 
Now  in  a  similar  manner,  with  like  restrictions,  we  extend  our 
notion  of  a  factor  still  farther  and  use  imaginary  numbers  as 
coefficients  or  as  terms  in  a  factor.  Consequently  a;^  + 1  may 
hereafter  be  regarded  as  factorable. 


IMAGIXARIES 


229 


For 

Similarly 
and 


x^  +  l  =  x^-  (-  1)  =  (x  +  V^)(x-  V^). 
4:i-2+9  =  4a;2-(-9)  =  (2a;  +  8V-l)(2x-3  V^) , 
a;2  +  6  =  a;2  -  (-  6)  =  (a;  +  V^)(x  -  V^). 

Further,  x^  —  1  =  (x  —  1)  (x"^  -{-  x  +1).  Hitherto  the  trinomial 
x^-{-x  +  l  has  been  regarded  as  prime  ;  but  the  student  can  easily 
prove  that  x'^  +  x  +  l  =  {x  +  ^  +  i  V^)  {x -\- ^  -  ^V—  3). 
Therefore  x^  —  1  has  three  factors,  ic  —  1,  x  +  -|-  +  |-  V—  3,  and 


^  +  i-iV-3. 

If  the  student  is  curious  as  to  the  way  in  which  the  factors 
oi  x^  -\-x-\-l  were  found,  he  may  discover  the  method  for  him- 
self by  studying  the  results  of  Exercise  7,  page  228. 

125.  Graphical  interpretation  of  pure  imaginaries.  In  our  pre- 
vious graphical  work  a  positive  number  and  a  numerically 
equal  negative  number,  as  +  2  and 
—  2,  were  represented  by  equal 
distances  measured  in  opposite 
directions,  such  as  OA  and  OB  of 
the  adjacent  figure.  Now  multi- 
plying .+  2  by  —  1  gives  —  2. 
Hence,  if  we  choose  to  do  so,  we 
may  regard  —  1  as  an  operator 
(rotor)  which  turns  OA  in  the 
direction  of  the  arrow  into  the 
position  OB,  or  through  two  right 
angles  (180  degrees). 

To  make  this  point  clearer  note  the  two  curves  of  the  figure 
on  page  230.  Curve  ABC  is  the  graph  of  the  function  x^  —x  —  2. 
Curve  A'B'C  is  the  graph  of  the  function  2  +  x  -  x\  The 
latter  function  was  obtained  by  multiplying  x"^  —  x  —  2  hj  —  1. 
The  graphical  effect  of  this  multiplication  is  to  turn  the  whole 
curve  ABC  about  X'X  as  an  axis  through  two  right  angles  to  the 
position  A'B'C'. 

The  preceding  illustrations  indicate  a  method  of  interpret- 
ing the  V—  1,  which  is  in  strict  conformity  with  our  previous 
graphical  work. 


230 


SECOND  COURSE  IN  ALGEBRA 


First  V— 1  •  V—  1  =  —  1.  Now  multiplying  a  number  by  —  1 
produces  the  same  effect  as  multiplying  twice  in  succession  by 

V— 1.  Therefore  multiply- 
ing by  V— 1  once  may  be 
regarded  as  producing  a 
rotation  of  one  right  angle 
(90  degrees),  or  one  half  as 
much  rotation  as  multiplying 
by-1. 

E-eturning  to  the  figure  on 
page  229,  OA,  or  +  2,  mul- 
tiplied by  V—  1  would  be 
turned  to  the  position  OK. 
Hence  the  point  K  is  said 
to  correspond  to  the  number 
2  V— 1.    Similarly,  point  R 


corresponds  to '  V— 1  and  point  G  to  3  V—  1.  And  OH  being 
measured  in  a  direction  opposite  to  OR,  OK,  and  00,  would 
correspond  to  —  V—  1.  In  like  manner  OL  corresponds  to 
—  2  V—  1.  This  last  result,  however,  may  be  reached  differ- 
ently. Multiplying  2  by  V—  1  three  times  gives  —  2  V—  1. 
These  successive  multiplications  by  V—  1  may  be  regarded 
as  producing  a  counterclockwise  rotation  through  three  right 
angles  which  would  locate  the  point  corresponding  to  —  2  V— 1 
on  0/  at  L  as  before. 

Therefore  the  graphical  representation  of  a,  pure  imafflm//-// 
number  b  V— 1  is  by  a  point  on  an  axis  perpendicular  (at  right 
angles)  to  the  axis  of  real  numbers,  b  units  in  the  direction  of 
OG  if  b  is  positive,  b  units  in  the  direction  of  01  if  b  is  nega- 
tive.   This  new  axis  will  be  called  the  imaginary  or  /-axis. 


TI 

H                   c 

.                                    ; 

t                                / 

T                               r 

t           ^w          2 

\          %l          t 

V^^  t     ^^J 

.X. ^g-1   -5Z^ v- 

^ ^^  0    m^i  ^ 

-           ^^^?           r- 

'              a           X 

r                   \ 

J.                     ^ 

I                     t 

f-                     -\ 

J '           ■                              vc' 

f' 

126.  Graphical  representation  of  a  complex  number.  The  com- 
plex number  ic  =  3  +  2  V— 1  consists  of  a  real  part  3  and  the 
imaginary  part  2  V— 1.  To  represent  such  a  number  we  meas- 
ure in  the  adjacent  figure  3  units  along  OX  from  0  to  R,  and 


IMAGINARIES 


231 


then  2  units  parallel  to  the  imaginary  axis  //'  from  R.  This 
gives  the  point  P,  which  is  the  graphical  representation  of  the 
complex  number 

cc  =  3  +  2  V^. 
If  the  student  pays  proper 
attention  to  signs,  he  should 
now  see  that  the  point  A  cor- 
responds to  ic'=  3  —  2  V  —  1, 
B  to  a;  =  — 2  +  4Vpi,  and 
Cto  cc  =— 4  —  V— 1. 

In  general,  if  x  is  a  com- 
plex number  a  +  hi,  x  is  rep- 
resented by  a  point  a  units 
from  the  imaginary  axis  and 
h  units  from  the  real  axis,  the  positive  and  negative  direction^ 
being  as  indicated  in  the  adjacent  figure. 

Note.  It  was  the  discovery  of  a  graphical  interpretation  for  the 
imaginary  numbers  which  did  more  than  anything  else  to  make  them 
mean  something  to  students  of  mathematics.  Until  this  discovery 
they  were  tolerated  because  their  appearance  as  the  roots  of  equa- 
tions was  a  constant  reminder  of  their  existence.  But  they  were 
usually  regarded  as  meaningless,  and  the  less  one  had  to  do  with 
them  the  better  he  liked  it.  About  1800  a  Norwegian  by  the  name 
of  Wessel,  and  the  Frenchman,  Argand,  gave  practically  the  same 
graphical  interpretation  as  that  found  in  the  text,  but  their  work 
was  little  noticed  till  Gauss  adopted  the  method  and,  by  his  influence 
and  ability,  placed  the  imaginary  number  on  a  firm  basis. 


I 

... 

B 

•3 -A 

rr 

'^- 

p 

, 

1 

\ 

w 

— 

-2    - 

0 

-\ 

0"' 

: 

I    \ 

^" 

c 

,^[ 

-f 

A 

i' 

Locate  the  point  'x,  if 


EXERCISES 


1. 

cc  =  2  +  I 

6. 

■r  =  -  3  -  2  I 

11. 

x  =  2±  V-3. 

2. 

X  =  3  +  3  /. 

7. 

X  =  4:  —   L 

12. 

x=S±3^-2 

3. 

x=^-2L 

8. 

X  =  4  +  4  i. 

13. 

ic  =  2H- V-11 

4. 

X  =  1  -  3  /. 

9. 

x=-l-l 

14. 

a;  =  2- V-11. 

5. 

x=-2+L 

10. 

X  =  -  3  -  5  /. 

15. 

cc  =  V-  12  -  5. 

232  SECOND  COURSE  IN  ALGEBRA 

The  graphical  interpretation  for  the  definition  of  equality  of  com- 
plex numbers  is  that  equal  complex  numbers  are  represented  by  the 
same  point  in  the  plane. 

The  student  should  not  conclude  from  the  preceding  exercises 
that  the  imaginary  axis  has  merely  replaced  the  y-axis  of  our  pre- 
vious graphical  work,  for  such  is  by  no  means  the  case.  We  have 
simply  used  our  device  of  rectangular  axes  to  represent  complex 
numbers.  The  fact  that  we  have  previously  made  use  of  the  same 
means  to  graph  a  function  should  not  embarrass  us  any  more  than  it 
would  disturb  us  to  play  croquet  on  a  baseball  ground.  The  fact  that 
both  take  place  on  the  same  field  does  not  make  them  the  same  game. 

For  some  jjurposes  it  is  convenient  to  relate  the  xy--p\&ne,  in  which 
real  number  relations  are  represented,  and  the  complex  plane  more 
closely  to  each  other.  In  fact,  we  may  proceed  as  if  we  had  merely 
added  a  third  axis,  the  axis  of  imaginary  numbers,  to  the  two  axes 
of  real  numbers  which  we  had  before.  This  axis  may  be  regarded 
as  passing  through  the  origin  at  right  angles  to  both  the  a:-axis 
and  the  y-axis.  By  means  of  it  we  may  locate  such  a  point  as  y  =  3, 
a:  =  2  -f  4  "v  —  1,  or  any  point,  in  fact,  in  which  one  coordinate  is  real 
and  the  other  imaginary  or  complex.  Then  we  may  go  on  and  con- 
struct the  imaginary  branches  of  many  of  those  curves,  the  real  graphs 
of  which  the  student  has  already  drawn. 

Note  on  use  of  imaginaries.  We  have  explained  the  laws  of  addi- 
tion, subtraction,  multiplication,  and  division  for  imaginary  (and 
complex)  numbers  and  have  made  some  use  of  them.  It  is  largely 
because  imaginaries  obey  these  laws  that  we  call  them  numbers,  for 
it  must  be  admitted  that  we  cannot  count  objects  with  imaginary 
numbers.  Nor  can  we  state  by  means  of  them  our  age,  our  weight, 
or  the  area  of  the  earth's  surface.  It  should  be  remembered,  how- 
ever, that  we  can  do  none  of  these  things  with  negative  numbers.  We 
may  have  a  grouj)  of  objects  —  books,  for  example  —  whose  number 
is  5 ;  but  no  group  of  objects  exists  whose  number  is  —  5,  or  —  3,  or 
any  negative  number  whatever.  If  it  be  asked,  How,  then,  can  neg- 
ative numbers  and  Imaginary  numbers  have  any  practical  use?  the 
answer  is  this :  They  have  a  practical  use  because  when  they  enter 
into  our  calculations  and  we  have  performed  the  necessary  opera- 
tions upon  them  and  obtained  our  final  result,  that  result  can  fre- 
quently be  interpreted  as  a  concrete  number  such  as  is  dealt  with 
in  ordinary  arithmetic.  Moreover,  if  the  result  cannot  be  so  inter- 
preted, it  is,  in  applied  mathematics  at  least,  finally  rejected. 

In  that  part  of  electrical  engineering  where  the  theory  and  meas- 
urement of  alternating  currents  of  electricity  are  treated,  complex 


IMAGINARIES  238 

numbers  have  had  extensive  use.  Their  employment  in  the  difficult 
problems  which  there  arise  has  given  a  briefer,  a  more  direct,  and  a 
more  general  treatment  than  the  earlier  ones  where  such  numbers 
are  not  used. 

In  theoretical  mathematics  complex  numbers  have  been  of*great 
value  in  many  ways.  For  example,  numerous  important  theorems 
about  functions  are  more  easily  proved  under  the  assumption  that 
the  variable  is  complex.  Then,  by  letting  the  imaginary  part  of  the 
complex  number  become  zero,  we  obtain  the  proof  of  the  theorem 
for  real  values  of  the  variable.  Indeed,  the  student  need  not  go  very, 
far  beyond  this  point  in  his  mathematical  work  to  learn  that,  if  e  is 
2.7128  +  (see  page  189),  e  "^-^  +  e  -"^-i  is  equal  to  the  real  number 
1.082  + .  At  the  same  time  he  will  learn  also  how  such  a  form  arises^ 
and  something  of  its  importance.  In  a  way  which  we  cannot  now 
explain,  even  so  involved  an  expression  as  (a  +  ih'y^^'^  has  in  higher 
work  a  meaning  and  a  use.  If  the  student  pursues  his  mathematical 
studies  far  enough,  that  meaning  and  use,  and  a  multitude  of  other 
uses  for  complex  numbers,  will  become  familiar  to  him.  But  the 
numbers  which  we  have  learned  in  this  book  to  use,  namely  fractions^ 
negative  numbers,  irrational  numbers,  and  complex  numbers,  com- 
plete the  number  system  of  ordinary  algebra,  for  it  can  be  proved 
that  from  the  fundamental  operations  no  other  forms  of  number 
can  arise. 


CHAPTER  XVII 

THEORY  OF  QUADRATIC  EQUATIONS 

127.  Character  of  the  roots  of  a  quadratic  equation.  It  is  often 
desirable  to  determine  the  character  of  the  roots  of  a  quadratic 
without  actually  solving  it.  To  determine  the  character  of  the 
roots  of  an  equation  means  to  find  out  whether  the  roots  are 
real  or  imaginary,  rational  or  irrational,  equal  or  unequal. 
These  properties  of  the  roots  of  a  given  quadratic  depend  on 
the  three  coefficients,  which  correspond  to  a,  h,  and  c,  in  the 
general  quadratic  equation  ax^  -\-  hx  +  c  =  0.  The  solution  of 
this  equation  gives  the  roots : 


r,  = 


and  1  .^  ,, 

This  expression  V^  —  A^ac  which  occurs  in  each  root  is  called 
the  discriminant  of  the  quadratic.  If  a,  b,  and  c  are  rational 
numbers,  it  is  evident  from  an  inspection  of  the  discriminant 
where  it  occurs  in  the  values  of  r^  and  r^,  that  the  following 
statements  are  true : 

I.  If  IP'  —  A  ac  is  -positive  and  not  a  perfect  square,  the  roots 
are  real,  unequal,  and  irrational. 

II.  If  IP  —  A  ac  is  positive  and  a  perfect  square,  the  roots  are 
real,  unequal,  and  rational. 

III.  If  IP"  —  A  ac  is  zero,  the  roots  are  equal. 

In  this  case  there  is  reallv  but  one  root, 

IV.  IflP  —  AiOcis  negative,  the  roots  are  imaginary. 

234 


^h-\-  ^fp  - 

-  4ac 

2(^ 

_  z,  _  V/>2  - 

-  4ac 

THEORY  OF  QUADRATIC  EQUATIONS  235 

EXERCISES 

Determine  the  character  of  the  roots  of  the  following  equa- 
tions by  the  use  of  the  discriminant : 

1.  2x^  +  5x-6  =  0. 

Solution:  h'^-4:ac  =  (5)2  _  4  •  2  •  (-  6)  =  25  +  48  =  73. 
Therefore  the  roots  are  real,  unequal,  and  irrational. 

2.  x^-5x  +  6  =  0.  8.  4aj2  =  9  -  9x. 

3.  5x^  -  llx  +  2  =  0.  9-  5a;  =  x^  +  5. 

10.  x^  -5x  +  7  =  0. 


4.  4.x' -20x-\- 25  =  0 

5.  5a;^ -3.^-3  =  0. 


11.  12x^-7x-\-6  =  0. 


12.  x(x-5)  =  x-16. 

6.  7£c'-2ic-hl0  =  0.        •  3       ^^ 

7.  x''  —  6  X  +  6  =  0.  ic''       X 

7        7x 

Determine  the  values  of  K  which  will  make  the  roots  of  the 
following  equations  equal.  (To  sa?y  the  roots  of  a  quadratic 
are  equal  is  the  usual  mathematical  way  of  stating  that  the 
equation  has  but  one  root.') 

15.  x''  -  Kx  +  16  =  0. 

Solution  :  a  =  1,  &  =  —  /i,  c  =  16. 

Hence  &2  _  4  f,^  =  k^  -  64. 

In  order  that  the  roots  be  equal,  h^  —  4:ac  must  equal  zero. 

Therefore  K'^  -  U  =  0. 

Whence  /v  =  ±  8. 

Check :  Substituting  8  for  K  in  the  original  equation, 

a,-2  -  8  X  +  16  =  0. 
Whence  a:  =  4,  only. 

Similarly,  substituting  /^  =  —  8,  x'^  +  8  a:  +  16  =  0. 
Whence  x  =  —  4,  only. 

16.  x^  -Kx  +  ^(j  =  0.  19.  x^  -10x  +  K  =  0. 

17.  x^  -3Kx-\-  81  =  0.  20.  2x^-\-Sx  +  K=0. 

18.  2x''-\-4.Kx  +  98  =  0.  21.  9x^ +  30x  + K +  9  =  0. 


236  SECOND  COURSE  IN  ALGEBRA 

22.  4:Kx' -60x^-25  =  0.       24.  ^9x'' -  (K -\-S)x +  A  =  0. 

23.  9K^x^-S4:X-^A9  =  0,     25.  (K^  +  5)x^-S0x  +  25  =  0. 

26.  (/^2  +  17)a;24-(5iiC-4)ic  +  4  =  0. 

Determine  the  relation  between  h  and  k  which  will  make 
the  roots  of  the  following  equations  equal : 

27.  A:V  +  Qhx  +  9  =  0.  29.  x^  -^4:kx  +  4^  =  0. 

28.  kx^  -2hx  +  16  =  0.       iSO.  kx^-2hx  +  6  =  0. 

Determine  the  values  of  a  for  which  the  following  systems 
will  have  two  sets  of  equal  roots  : 

31.^'  =  ^',       ■        y.3i.'''  +  '-^=/' 

y  =  x-\-l.  y  =  x-{-l. 

f  =  2x,  ^,    x^-{-y^  =  2x, 

f  32.  -^  /  34.  -^  .  ' 

y  =  x  -\-  a.  y  =  X  -\-  a. 

128.  Relations  between  the  roots  and  the  coefficients.    The  roots 
of  ax^  +  &c  +  c  =  0  are  

(1) 
2a  (2) 

(1)  +  (2)  gives  r,  +  r,  =  ^  =  -  ~  (3) 

(1)  X  (2)  gives       V,  =  *!:z(|^i^  =  £.  (4) 

The  general  quadratic  equation  may  be  written 

x^^tx  +  -  =  0.  (5) 

a ~€b 

Then  for  any  quadratic  in  which  the  coefficient  of  a;^  is  1 : 

I.  From  (3)  and  (5)  the  coefficient  of  x,  -  5  is  the  sum  of  the 
roots  with  the  sign  changed. 

II.  From  (4)  and  (5)  the  constant  term,  -  j  is  the  product  of 
the  roots. 

I  and  II  may  be  used  to  form  a  quadratic  whose  roots  are  given. 


-h  + 

V//^- 

-  4:ac 

2a 

-h- 

-s/b^- 

-  4ac 

THEORY  OF  QUADRATIC  EQUATIONS     237 

EXERCISES 
Form  the  quadratic  whose  roots  are : 

1.  5,-3. 

Solution :  r^  +  r^  with  the  sign  changed  =  —  2. 
r-j  X  Tg  =  5  (—  3)  =  —  15. 
Hence  the  required  equation  isx^  —  2x  —  15  =  0. 

2.  2,  7.  9.  -  3  ±  Vs.  15.  a,  c. 

3.-3,10.  10.  f±V7.  _        1 

A         A        K  r  16.  a,-- 

4.-4,-5.  •  11.  |±  ^Ve.  « 

5.  -12,-1.  -6±2V3  2^^ 

^6.  1,5.  ^^'  5  ^^-  ^'''    3 

7.  10,  - 1.  13.  V5,  -  3  Vs.  1 

8.  2+V3,2-V3.  14.  3-V2,2+V2.  ^®-''+^'^3i* 

Solve  the  following  equations  and  check  each  by  showing 
that  the  sum  of  the  roots  with  its  sign  changed  is  the  coefficient 
of  X,  and  that  the  product  of  the  roots  is  the  constant  term : 

19.  a;2  -  12a:  -  13  =  0.  21.  cc^  +  3ic  +  3  =  0. 

20.  x"  -  10a;  +  16  =  0.  22.  a;'  -  5a;  +  20  =  0. 

23.  a;'  +  2ic  +  2  =  0. 

24.  One  root  of  cc^  —  4  a?  — 12  =  0  is  —  2.   Find  the  other  root. 
Sohition  :  Let  r^  be  the  required  root. 

Then  -  (r^  +  r^)  =  -  (-  2  +  r2)  =  -  4. 

Solving,  ^2  =  ^• 

Check :  r^r^  =  (-  2)  (6)  =  -  12. 

25.  One  root  of  a;^  +  7  a;  — 18  =  0  is  —  9.    Find  the  other  root. 
Find  the  value  of  the  literal  coefficient  in  the  following : 

26.  a;^  +  2  a;  —  c  =  0,  if  one  root  is  3. 

27.  a;^  —  a;  —  c  =  0,  if  one  root  is  10. 

28.  a;^  +  8  a;  —  c  =  0,  if  one  root  is  —  2. 

29.  x^  —  ex  —  70  =  0,  if  one  root  is  10. 

30.  x^  +  2  ^'a;  +  25  =  0,  if  one  root  is  -  5. 


238  SECOND  COURSE  IN  ALGEBRA 

31.  05^  —  3  ax  —  62  =  0,  if  one  root  is  4. 

32.  2x^  —  llx  +  c  =  0,  if  one  root  is  5. 

33.  a*2  -  20x  + 12  =  0,  if  one  root  is  f. 

34.  ax^  —  6  cc  —  21  =  0,  if  one  root  is  —  3. 

35.  x^  —  Sx  -]-  c  =  0,  if  one  root  is  three  times  the  other. 

36.  x"^  +  7  X  +  c  =  0,  it  one  root  exceeds  the  other  by  1. 

37.  x"^  -\- 11 X  ■}- b  =  0,  if  the  difference  between  the  roots  is  9. 

38.  x^  —  5  X  —  c  =  0,  ii  the  difference  between  the  roots  is  7. 

39.  x^  —  5  X  —  a  =  0,  it  the  difference  between  the  roots  is 
-13. 

129.  Number  of  roots  of  a  quadratic.   Up  to  this  we  have  assumed 

that  a  quadratic  equation  has  but  two  roots.    This  fact  can  be  proved 

from  the  preceding  work  as  follows  : 

If  we  write  the  equation  ax^  +  hx-\-  c  =Oin  the  form  x^  +  -x  +  -  =  0 

a        a 

and  substitute  therein  from  (3)  and  (4)  on  page  2.36,  we  get  x^—  (r^  ■{■r^)x-\- 
r^r^  =  0.  This  can  be  factored  and  written  as  (x  —  r^)  (x  —  r^)  =  0. 
Now  if  any  value  of  x  different  from  r^  and  rg,  say  r^,  be  a  root  of 
this  equation,  such  a  value  when  substituted  for  x  must  satisfy  the 
equation  (x  —  r^)  (x  —  r^)  =  0. 

Hence  (r^  —  r^)  (r^  —  r^)  must  equal  zero.  By  definition,  how- 
ever, Tg  is  different  from  r^  and  r^.  Consequently  neither  the  factor 
(rg  —  rj)  nor  (r^  —  r^)  can  equal  zero,  and  therefore  their  product  can- 
not equal  zero.  This  proves  that  no  additional  value,  r^,  can  satisfy 
the  equation  x^  —  (r^  -|-  rg)^  -|-  r^r^  =  0.  As  this  equation  is  but  another 
form  of  ax^  -{■  bx  -\-  c  =  0,  the  latter  has  only  two  roots. 

130.  Formation  of  equations  with  given  roots.  The  method  of 
forming  quadratic  equations  which  was  used  in  the  preceding 
exercise  applies  to  equations  having  two  roots  only.  A  reversal 
of  the  method  of  solving  equations  by  factoring  (page  31), 
however,  enables  us  to  build  up  an  equation  with  any  num- 
ber of  given  roots. 

The  correctness  of  the  method  for  three  given  roots  will  be 
clear  from  what  follows  : 

Form  the  equation  whose  roots  are  a,  b,  and  c. 

Write  (x  -a)(x-  b)  (x  -  c)  =  0.  (1) 


THEORY  OF  QUADRATIC  EQUATIONS  239 

Now  if  a  is  put  for  x  in  (1),  we  obtain  0  =  0. 

Similarly,  if  ^  or  c  is  put  for  x  in  (1),  the  result  is  0  =  0. 

Therefore  (1)  is  the  equation  whose  roots  are  a,  b,  and  c,  and 
the  expanded  form  of  (V),x^  —  (a-\-h  -\-  c)  x^  +  (ab  -{-ac-\-hc)x-\- 
abc  =  0,  is  the  required  equation. 

The  same  reasoning  applies  if  we  form  in  this  way  an  equar- 
tion  with  any  number  of  given  roots. 

Note.  The  relations  between  the  roots  and  the  coefficients  of  an 
equation  were  discovered  at  about  the  same  time  by  Vieta  in  France, 
by  Girard  in  Holland,  and  by  Harriot  in  England.  Vieta  actually 
wrote  the  cubic  equation  in  about  the  form  given  in  the  text  so 
as  to  display  these  relations.  The  very  important  algebraical  theo- 
ries which  result  from  these  properties  were  developed  in  detail  by 
Newton,  and  have  been  the  subject  of  study  by  many  of  the  most 
distinguished  mathematicians  since  his  time. 

EXAMPLES 

1.  Form  the  equation  whose  roots  are  3  and  —  5. 

Solution  :  By  the  conditions,    x  =  3  and  x  =  —  6. 

Therefore  x  —  3  =  0     and     a:  +  5  =  0. 

Then  (x  -  3)  (x  +  5)  =  0.  (1) 

Expanding,  x^ -\- 2  x -15  =  0.  (2) 

Substitution  shows  that  3  and  —  5  are  the  roots  of  (1)  and  (2). 

2.  Form  the  equation  whose  roots  are  1,  3,  and  —  2. 

Solution  :  As  before,  x  —  1  =  0,  a;  —  3=0,  and  a;  +  2  =  0. 

Therefore  (x  -  1)  (x  -3)  (x  +  2)  =  0.  (1) 

Expanding,  x^  -  2  x^  -  5  x  +  Q  =  0.  (2) 

Inspection  shows  that  the  given  roots  1,  3,  and  —  2  satisfy  the 
equations  (1)  and  (2). 

EXERCISES 

By  the  method  used  in  the  preceding  examples  form  the 
equation  whose  roots  are : 

1.  3,  7.  4.  2  ±  V5.  6.  3,  -  3,  8. 

2.  4,-5,6.  3±V7  7.  1,  |,  -  2. 

3.  1  +  V3,  1  -  V3.  2       '  8.  1  ±  V3,  3. 


9. 

a  -\-  b,  a  — 

10. 

1       K 

-i  oa. 
a 

11. 

3c±  V2a. 

io 

4a±V3c 

240  SECOND  COURSE  IN  ALGEBRA 

h.  13.  i\,  r^  1-3. 

14.  3,  2  ±  -v^. 
15.-5,  -7,6,  8. 

16.  2  ±  V3,  3  ±  V2. 

17.  1,  -2,a±  Va. 

131.  Factors  of  quadratic  expressions.    Let  r^  and  r^  be  the 

roots  of  ax^  -\-  bx  -{-  c  =  0. 

b  G 

Then  x^  -j-  -  x-\ —  =  a;^  ~  (^1  +  ^2)  ^  "^  ^1^2  =  (^  —  ^j)  (^  — 1\^, 

or  ax'^  +  &a;  +  c  =  a  (rr  —  r^)  (ic  —  r^. 

Therefore  the  three  factors  a,  x  —  r^,  and  x  —  r^  of  any 
quadratic  expression  can  be  written  if  we  first  set  the  expres- 
sion equal  to  zero  (see  §  17)  and  solve  for  r^  and  r^  the  equation 
thus  formed.  Obviously  the  character  of  the  roots  so  obtained 
will  determine  the  character  of  the  factors.  K*iin>S6"t.^  Lhe  use  of 
the  discriminant  P  —  Aac  we  can  decide  whether  the  factors  of  a 
quadratic  expression  are  real  or  imaginary,  rational  or  irrational, 
without  factoring  it. 

EXERCISES 

Determine  which  of  the  following  expressions  have  rational 
factors  : 

1.  j^2  _  3^  _  4Q  e.  5x^  +  Sx-  20. 

2.  2x^-\-5x-7.  7.  3a;2-9ic  +  28. 

3.  7cc2-9a^-f-18.  8.  SS  h'' -  2SS h  -  6. 

4.  24:x^  -  X  -  10.  9.  x^  -2ax-h  (a^  -  b^). 

5.  72  a;2  -  17  ic  +  1.  10.  abx^  -  (b^  -^  a"")  x -{■  ab. 
Separate  into  rational,  irrational,  or  imaginary  factors  : 

11.  2cc2_^5a!-8. 

Solution :  Let  2  a;^  +  5  a;  -  8  =  0. 

Solving  by  formula,  x  = ^ ^^ ^  =  — '- — 


THE  BINOMIAL  THEOREM  243 

The  preceding  expansion  expresses  in  symbols  the  law  known 
as  the  binomial  theorem.  The  theorem  holds  for  all  positive 
values  of  n  and  with  certain  limitations  (see  §  136)  for  nega- 
tive values  as  well.    This  will  be  assumed  without  proof. 

Note.  The  coefficients  of  the  various  terms  in  the  binomial  expan- 
sion are  displayed  in  a  most  elegant  form  as  follows : 

1/ 

1^  3  1 
1-4641 

In  this  arrangement  each  row  is  derived  from  the  one  above  it  by 
observing  that  each  number  is  equal  to  the  sum  of  the  two  num- 
bers, one  to  the  right  and  the  other  to  the  left  of  it,  in  the  line  above. 
Thus  4  =  1+3,  6  =  3-1-3,  etc.  The  next  line  is  1  5  10  10  5  1. 
The  successive  lines  of  this-  table  give  the  coefficients  for  the  expan- 
sions of  (a  +  by  for  the  various  values  of  n.  Thus  the  numbers  in 
the  last  line  of  the  triangle  are  seen  to  be  the  coefficients  when  w  =  4 ; 
the  next  line  would  give  those  for  n  =  5.  This  arrangement  is  known 
as  Pascal's  Triangle,  and  was  published  in  1665.  It  was  probably 
known  to  Tartaglia  nearly  a  hundred  years  before  its  discovery  by 
Pascal. 

134.  The  factorial  notation.  The  notation  5!,  or  [5,  signifies 
1- 2- 3 -4.  5,  or  120.    Similarly,  4 !  =  1  •  2 •  3 •  4  =  24. 

In  general,  nl  =  1  -  2  -  S ■  4t  -  ■  ■  (n  —  2)(n  —  l)n. 

The  sign  nl,  or  [?z,  is  read  factorial  n. 

With  the  factorial  notation  the  denominators  of  the  third 
and  fourth  terms  in  the  expansion  of  (a  +  by  in  §  133  become 
2 !  and  3 !  respectively. 

EXERCISES 

Expand  by  the  rule  : 

1.  (a  +  bf.  3.  (a  +  ly.  5.  (a  +  3)1 

2.  (a  -  ly.  4.  (a  +  2)«.  6.  (2  -  a)\ 
Obtain  the  first  four  terms  of : 

7.  (a  +  by^       8.  (a  +  b^".       9.  (a  +  1)^«.        10.  (a  -  2^". 


12 


244  SECOND  COURSE  IN  ALGEBRA 

Expand : 

11.  (a*  -|-  2  by.  Hint.  To  avoid  confusion  of  exponents  first  write 
(a^y  +  (ay  (2  by  +  (ay  (2  by  +  (ay  (2  by  + 

(ay  (2  by  ■{■(2  by. 

Then  in  the  spaces  left  for  them  put  in  the  coefficients  according 
to  the  rule  of  §  133. 

Finally,  expand,  and  simplify  each  term. 

13.  (a^-{-2by.  \         ^J  \         «/ 

Obtain  in  simplest  form  the  first  four  terms  of : 
16.  (a2  +  2^.)2«.  20.  (a^-Sby^. 

■'■(•■-!)"  "(7-^)' 

''{7 -A}     -(f-^r- 

24.  Write  the  first  six  terms  of  the  expansion  of  (a  +  by  and 
test  it  for  n  =  1,  n  =  2,  n  =  S,  and  n  =  4c.  How  does  the  num- 
ber of  terms  compare  with  n?  What  is  the  value  of  each 
coefficient  after  the  (n  -f  l)st  ?  Why  does  not  the  expansion 
extend  to  more  than  six  terms  when  n  =  5? 

Compute  the  following,  correct  to  two  decimal  places : 

26.  (1.1)^°.    Hint.  (1.1)1° ,,  (i  +  j>)W  etc.  28.  (2.9)1 

27.  (.9Sy\    Hint.  (.98)1i  =  (1  -  .02)",  etc.  29.  (1.06)^ 

30.  6!.  1111 

31.  2! -4!.  ^^'^~^2l'^Si'^H~^5l' 

32.  6!  ^3!.  1514131211109 

33.  4! -3!- 2!  2!.  7! 


25.  Write  the  first  four  terms 


THE  BINOMIAL  THEOREM  245 

135.  Extraction  of  roots  by  the  binomial  theorem.  By  refer- 
ence to  the  expansion  (1),  page  242,  it  can  be  seen  that  none 
of  the  factors  n,  n  —  1,  n  —  2,  n  —  S,  etc.  become  zero  for  frac- 
tional or  negative  values  of  n.  Hence  for  such  exponents  the 
development  of  (a  -f  by  becomes  an  infinite  series.  If  a  is 
numerically  greater  than  b,  and  n  has  any  one  of  the  values 
h  h  h  6tc.,  the  resulting  series  has  a  limiting  value.  In  those 
expansions  where  a  is  considerably  greater  than  b,  this  value 
can  be  readily  approximated  by  finding  the  sum  of  the  first 
few  terms.  Therefore  the  square  root,  cube  root,  and  all  other 
roots  can  be  obtained  approximately  by  the  aid  of  the  binomial 
theorem. 

Note.  The-process  of  extracting  the  square  root  and  even  the  cube 
root  by  means  of  the  binomial  expansion  was  familiar  to  the  Hindus 
more  than  a  thousand  years  ago.  The  German,  Stifel  (1486-1567), 
stated  the  binomial  theorem  for  all  powers  up  to  the  seventeenth,  and 
also  extracted  roots  of  numbers  by  this  method. 

EXAMPLES 

Find  to  three  decimals  by  the  binomial  theorem : 

1.  (27)i 

Solution :  (27)^  =  (25  4-  2)^         >.  v 

,     =262^1.  25-^.  2)- 1.25-^.22 +3-1^.25-^.23... 

=  5 +  .2 -.001 +  .00016  =  5.196+. 

2.  (67)3. 

Solution:   (67)^  =  (64  +  3)^ 

=  64^  +  1  .  64- 1 . 3  -  1  .  64- 1 .  32  +  -g\  .  64"  t .  33  . . . 


I 


T^        T02T 

4  +  .0625  -  .00097  =  4.0615. 


Here  three  terms  give  the  result  correct  to  five  figures. 

3.  (79)i 

Hint.  (79)^  =  (81  -  2)2=  81^  -  1  ■  81"^  •  2  +  i  •  SI"!  .22  +  .... 

Here  (81  —  2)^  yields  more  accurate  results  with  fewer  terms 
than  does  (64  +  15)^. 


246  SECOND  COURSE  IN  ALGEBRA 

EXERCISES 

Find  to  two  decimals  by  the  binomial  theorem : 

1.  (26)i  3.  (79)1  5.  (28)i  7.  (25)i 

2.  (38)^.  4.  (120)i  6.  (6Q)K  8.  (700)1 

Find  the  first  four  terms  of  :  / 

9.  (l  +  a;)i  11.  (3-a:)i  IS.  (2  +  x)K  ^ 

10.  (2  +  x)^.  12.  (l-\-x)i  14.  (S-x)K 

136.  Limitations  on  a  and  &  in  (a  +  b)".  The  expansion 
(a  +  by  has  a  meaning  for  all  values  of  n,  only  if  a  and  ^  are 
properly  chosen.  To  illustrate  the  truth 'of  this  statement  we 
shall  consider  the  expansion  of  (l  +  ic)"^  for  various  values 
of  X.    By  the  theorem, 

(1  +  x)-'^  =  1  _aj  +  cc2  _  ^3  ^  ^.4  _  ^5  _|_  ^6 _        ^^^ 

Now  (1  +  x)~^  =  -j Hence  the  left  member  of  (1)  has 

a  meaning  for  all  values  of  x  except  —  1.  The  right  member 
of  (1)  is  an  infinite  geometrical  series  whose  ratio  is  —  x.  This 
series  has  a  limiting  value  only  when  x  is  numerically  less 
than  1  (see  page  173);  that  is,  if  x  be  a  positive  or  negative 
proper  fraction.  For  positive  or  negative  values  of  x  numeri- 
cally greater  than  1  the  series  has  no  definite  value.  There- 
fore the  expansion  has  a  meaning  only  when  x  is  numerically 
less  than  1.  Here  1  corresponds  to  a  and  x  to  b  in  (a  -\-  b)% 
and  the  preceding  discussion  indicates  but  does  not  prove  the 
truth  of  the  following  statement : 

The  expansion  (a  -\-  b)^  has  a  definite  value  if  n  is  positive  or 
negative,  integral  or  fractional,  provided  a  is  greater  than  b. 

A  proof  of  this  last  statement  is  beyond  the  scope  of  this 
book. 

Ifote.  The  binomial  theorem  occupies  a  remarkable  place  in  the 
history  of  mathematics.  By  means  of  it  Napier  \^as  led  to  the  dis- 
covery of  logarithms,  and  its  use  was  of  the  greatest  assistance  to 


THE  BINOMIAL  THEOREM  247 

Newton  in  making  his  most  wonderful  mathematical  discoveries. 
But  to-day  the  results  of  Napier  and  of  Newton  are  explained  with- 
out even  so  much  as  a  mention  of  the  binomial  theorem,  for  simpler 
methods  of  obtaining  these  results  have  been  discovered. 

It  was  Newton  who  first  recognized  the  truth  of  the  theorem,  not 
only  for  the  case  where  n  is  a  positive  integer,  which  had  long  been 
familiar,  but  for  fractional  and  negative  values  as  well.  He  did  not 
give  a  demonstration  of  the  general  validity  of  the  binomial  develop- 
ment, and  none  even  passably  satisfactory  was  given  until  that  of 
Euler  (1707-1783).  The  first  entirely  satisfactory  proof  of  this  diffi- 
cult theorem  was  given  by  the  brilliant  young  Norwegian,  Abel 
(1802-1829). 

137.  The  rth  term  of  (a-\-b)".  According  to  the  binomial 
theorem  the  fifth  term  of  the  expansion  (1)  on  page  242  is 

%i(n  -  l)(n  -2)(n  -3)a»-^b'' 

l-^.^-4: 

If  we  note  particularly  this  term  and  those  on  page  242,  we 
can  write  down,  from  the  considerations  which  follow,  any 
required  term  without  writing  other  terms  of  the  expansion. 

The  denominator  of  the  coefficient  of  the  fifth  term  is  4!. 
From  the  law  of  formation  the  denominator  in  the  sixth  term 
would  be  5!,  in  the  seventh  term  6!,  etc.  Consequently  in  the 
rth  term  the  denominator  of  the  coefficient  would  be  (r  —  1) !. 

The  numerator  of  the  coefficient  of  the  fifth  term  contains 
the  product  of  the  four  factors  n(n  —  l){n  —  2)(n  —  3).  The 
sixth  term  would  contain  these  four  and  the  factor  n  —  4:. 
Similarly,  the  last  factor  in  the  seventh  term  would  be  n  —  5, 
etc.  Hence  the  last  factor  in  the  rth  term  would  he  n  —  (r  —  2), 
Therefore  the  numerator  of  the  coefl&cient  of  the  rth  term  is 
n(n  -l)(n-  2)(n  -  3)  ■  -  •  (n  -  r  +  2). 

The  exponent  of  a  in  the  fifth  term  is  ti  —  4,  in  the  sixth 
term  it  would  be  n;  —  5,  etc.  Therefore  in  the  rth  term  the 
exponent  of  a  would  be  n  —(r  —  1)  or  n  —  r  -\-l. 

The  exponent  of  h  in  the  fifth  term  is  4,  in  the  sixth  term  it 
would  be  5,  etc.  Therefore  in  the  rth  term  the  exponent  of 
h  would  be  r  —  1. 


248  SECOND  COURSE  IN  ALGEBRA 

The  sign  of  any  term  of  the  expansion  (if  71  is  a  positive 

integer)  is  plus  if  the  binomial  is  a-\-h.    If  the  binomial  is 

a  —  b,  the  terms  containing  the  odd  powers  of  h  will  be  nega^ 

tive ;  the  sign  of  the  rth  term  being  minus  if  r  —  1  is  odd. 

Therefore  the  rth  term  (r  ^  1)  of  (a  +  by  equals  plus  or  minus 

n(n-l)(n-2)(n-3)...(n-r+2)    ^_^^      _, 

(r-l)I  •     ^^ 

The  formula  for  the  (r  +  l)st  term  is  more  simple  and  more 
easily  applied.    It  is  plus  or  minus 

n{n-l){n-2){n-Z).-.{n-r  +  l)_^,^^_  (2) 

If  we  wanted  the  12th  term,  we  would  in  using  (1)  substi- 
tute 12  for  r,  and  in  using  (2)  we  would  substitute  13  for  r.'\ 

*  • 

EXERCISES 

Write  the :  10'^^'^^ 

1.  5th  term  of  {a  +  bf. 

Solution :  Substituting  10  for  n  and  5  for  r  in  the  formula  (1)  gives 

4!  4.3.2 

2.  6th  term  of  (a  +  b)\  ^    a^.   ^  .  (a      PV' 

^  ^  8.  6th  term  of    7 • 

3.  4th  term  of  (a  +  by".  V       «/ 

4.  7th  term  of  (a  -  b)-  9    ^,^  ,,,,^  ,f  /^  _  2^Y', 

5.  8th  term  of  (a  -  Z')^^  \^        ^ 


10.  middle  term  of  (x^  —  x^*^. 

11.  Sthtermof  (  V^->J^j  • 


/         1\^ 

6.  4th  term  of  I  a  +  -  |  • 

7.  5th  term  of  (a^  -  bf\ 

Find  the  coefficient  of : 

12.  x''  in  (1  +  xy\         '  14.  x'''  in  (ic»  +  1)^^ 

13.  x^  in  (1  +  xy\    '  15.  x^°  in  (x^  -  x-y\ 

16.  Expand  (3  +  1)-^  and  (1  +  3)-^  by  the  binomial  theorem, 
and,  if  possible,  find  the  sum  of  each  series  thus  obtained. 

17.  Treat  (2  +  l)-^  and  (1  +  2)-^  as  in  Exercise  16. 


^ 


CHAPTER  XIX 

SUPPLEMENTARY  TOPICS 

138.  Mathematical  induction.  The  truth  of  many  theorems 
which  may  be  stated  as  formulas  involving  a  certain  letter 
can,  for  integral  values  of  this  letter,  be  established  with  cer- 
tainty and  elegance  by  a  method  of  proof  known  as  mathe- 
matical induction.  This  process  will  best  be  understood  by 
an  illustrative 

Example :  Let  us  suppose  that  some  one  has  discovered  the 
remarkable  relation  that  the  sum  of  the  cubes  of  the  consecutive 
integers  1,  2,  3,  •  •  •,  n  is  equal  to  the  square  of  the  sum,  of  the 
integers.  Further,  let  us  assume  that  he  had  expressed  this  in 
the  formula 

13  +  23  +  3^  +  .  • .  +  71^  =  (1  +  2  +  3  +  •  •  •  -]-  nf.         (1) 

Lastly,  let  us  suppose  that  he  has  tested  the  truth  of  this  for- 
mula for  all  the  positive  integral  values  of  n  up  to  15  and  has 
become  convinced  that  the  formula  is  true  for  any  positive  inte- 
gral value  of  n.  He  knows,  however,  that  he  has  not  proved  the 
formula  for  n  =  lQ,  nor  for  any  one  of  the  infinite  number  of 
integers  greater  than  15.  The  labor  of  verifying  by  actual 
substitution,  however,  has  become  too  great  and  he  desires  a 
general  proof  of  (1). 

He  could  arrive  at  such  a  proof  in  the  following  manner : 
The  right  member  of   (1)   is   an  arithmetical  series  of   n 
terms ;  the  first  term  is  1  and  the  common  difference  is  1. 

Substituting   in   S  =  -\_2  a  -\-  {n  —  1)  d'\  gives    S  =     ^      — ^  • 

Therefore,  if  k  replaces  /i,  (1)  may  be  written 

13-1-23-1-33-f  ..•-f-F  =  (l  +  2-f-3  +  ...  +  A-)^  =  r^^^^^l-  (2) 

249 


248  SECOND  COURSE  IN  ALGEBRA 

The  sign  of  any  term  of  the  expansion  (if  ti  is  a  positive 

integer)  is  plus  if  the  binomial  is  a-\-h.    If  the  binomial  is 

a  —  h,  the  terms  containing  the  odd  powers  of  b  will  be  nega^ 

tive ;  the  sign  of  the  ?'th  term  being  minus  if  r  —  1  is  odd. 

Therefore  the  rth  term  (r  ^  1)  of  (a  +  ly  equals  plus  or  minus 

n(n~l)(7z-2)(n-3)...(n-r+2)     ^^^      _, 

(r-1)!  *     ^  ^ 

The  formula  for  the  (r  +  l)st  term  is  more  simple  and  more 
easily  applied.    It  is  plus  or  minus 

»(»-l)(»-2)(»-3)...(n-r  +  l)^„.,^,_  ^2) 

If  we  wanted  the  12th  term,  we  would  in  using  (1)  substi- 
tute 12  for  r,  and  in  using  (2)  we  would  substitute  13  for  r.' f 

EXERCISES 

Write  the:  .^_  ^^^^ 

1.  5th  term  of  {a  +  hf^ 
Solution :  Substituting  10  for  n  and  5  for  r  in  the  formula  (1)  gives 

MLi-ill  aV  =  i^^  a%'  =  210  a«6«. 

2.6thter„>of(a  +  .)».  3.  eth  term  of  (^  -  ^Y' 

3.  4th  term  of  {a  +  bf.  \l>       "-I 

4.  7th  term  of  (a  -  lif.  9    7^^  term  of  f^-  -  ^'Y'. 


5.  8th  term  of  {a  -  hy\ 

I         1\^ 

6.  4th  term  of  I  a  +  -  j 


10.  middle  term  of  {x"  -  xf. 


11.  5th  term  of 


(^-^r 


7.  5th  term  of  {a^  -  hf\ 
Find  the  coefficient  of : 

12.  x^  in  (1  +  xy\  14.  a;i5  in  (ic«  +  1)''. 

13.  £c8  in  (1  +  a^')'*.    '  15.  x^""  in  {x^  -  x-y\ 

16.  Expand  (3  -f  1)-^  and  (1  +  3)-^  by  the  binomial  theorem, 
and,  if  possible,  find  the  sum  of  each  series  thus  obtained. 

17.  Treat  (2  +  1)"^  and  (1  +  2)-^  as  in  Exercise  16. 


a 


CHAPTER  XIX 

SUPPLEMENTARY  TOPICS 

138.  Mathematical  induction.  The  truth  of  many  theorems 
which  may  be  stated  as  formulas  involving  a  certain  letter 
can,  for  integral  values  of  this  letter,  be  established  with  cer- 
tainty and  elegance  by  a  method  of  proof  known  as  mathe- 
matical induction.  This  process  will  best  be  understood  by 
an  illustrative 

Example :  Let  us  suppose  that  some  one  has  discovered  the 
remarkable  relation  that  the  sum  of  the  cubes  of  the  consecutive 
integers  1,2,  3,  ■•  -,  n  is  equal  to  the  square  of  the  sum  of  the 
integers.  Further,  let  us  assume  that  he  had  expressed  this  in 
the  formula 

13  +  23  +  3«  +  •  •  •  +  7.^  =  (1  +  2  +  3  +  . .  ■  +  n)\         (1) 

Lastly,  let  us  suppose  that  he  has  tested  the  truth  of  this  for- 
mula for  all  the  positive  integral  values  of  n  up  to  15  and  has 
become  convinced  that  the  formula  is  true  for  any  positive  inte- 
gral value  of  n.  He  knows,  however,  that  he'  has  not  proved  the 
formula  for  ti  =  16,  nor  for  any  one  of  the  infinite  number  of 
integers  greater  than  15.  The  labor  of  verifying  by  actual 
substitution,  however,  has  become  too  great  and  he  desires  a 
general  proof  of  (1). 

He  could  arrive  at  such  a  proof  in  the  following  manner : 
The  right  member  of   (1)  is  an  arithmetical  series  of   n 
terms ;  the  first  term  is  1  and  the  common  difference  is  1. 

Substituting   in   S  = -\2  a -\-  (n  —  V)  d'\  gives    S  =  — ^^ — - — —  • 

Therefore,  if  Iz  replaces  n,  (1)  may  be  written 

l«-f2^+3^+---  +  /^^  =  (l  +  2+3  +  -..  +  /0^  =  r^i^^P  (2) 

249 


250  SECOND  COURSE  IN  ALGEBRA 

Now  (2)  is  known  to  be  true  for  any  value  of  k  between 
1  and  15.    Consequently  we  can  obtain  the  correct  formula  for 
k  -\-ly  or  16,  terms  by  actually  adding  the  (k  +  l)st  term  to 
each  member.    This  gives 
1«  +  2«  +  3«  +  •  • .  +  A;«  H-  (A^  + 1)«  =  ^"^(J^^^y  _f_  (7,  _^  1)8       (3>) 

=  (^^  +  l)^(j  +  /.  +  l)        (4) 

Now  (2)  is  the  correct  formula  for  all  the  numbers  tried, 
while  (6)  is  the  correct  formula  for  all  the  numbers  tried  and 
for  one  not  tried,  the  number  16.  Another  important  fact 
about  (6)  is  that,  though  derived  through  equations  (3),  (4), 
and  (5),  it  can  also  be  obtained  by  merely  substituting  k  +  1 
for  k  in  (2).  This  is  obvious  on  inspection.  We  have  proved, 
then,  that  (1)  is  true  for  any  integral  value  of  n  from  1  to  16, 
but  without  actually  substituting  16  for  n.  This,  however,  is 
equivalent  to  proving  that  if  (1)  is  true  for  n  terms,  it  is  true 
for  71  +  1  terms.  But  (1)  is  true  for  15  terms ;  therefore  it  is 
true  for  16  terms.  'Again,  if  true  for  16  terms,  (1)  is  true  for 
17  terms,  etc.  Therefore  (1)  is  true  for  any  positive  integral 
value  of  n. 

The  steps  in  a  proof  by  mathematical  induction  are  : 

1.  Test  the  given  or  derived  formula  for  integral  values  of  n 
until  its  general  truth  seems  probable. 

2.  Write  the  formula  with  k  in  place  of  n,  k  denoting  any 
number  within  the  range  of  trial. 

3.  Derive  (or  establish  the  truth  of)  the  correct  formula  for 
a  value  of  k  greater  by  1.  (In  the  case  of  a  series  this  can 
frequently  be  done  by  actually  adding  the  (A;  +  l)st  term  to 
both  members  of  the  equation ;  for  many  formulas  some  other 
appropriate  device  must  be  thought  out.)   Then,  if  the  formula 


SUPPLEMENTARY  TOPICS  251 

thus  obtained  is  the  same  as,  or  can  be  reduced  to,  the  original 
one,  with  k  -\-l  put  for  k,  we  have  established  the  truth  of  the 
formula  for  one  integer  beyond  the  range  of  trial. 

4.  Reason  thus  :  If  the  formula  is  true  for  n  =  k,  it  is  true 
for  n  =  k  -\-l.  But  it  is  true  (let  us  say)  for  k  =  5;  therefore 
it  is  true  for  k  —  6.  But  if  true  for  A:  =  6,  it  is  true  for  k  =  1, 
and  so  on.  Therefore  the  formula  is  true  for  all  positive  inte- 
gral values  of  n. 

Observe  that  a  proof  of  a  formula  or  theorem  by  mathematical 
induction  is  possible  only  for  integral  values  of  n.  (A  series,  for 
example,  cannot  contain  3i  terms.) 

Frequently  a  formula  is  true  only  when  n  is  even  or  when  n  is  odd. 
In  such  cases  the  next  higher  value  of  n  is  greater  by  2. 

Note.  It  is  astonishing  how  far  certain  formulas  will  stand  numer- 
ical verification,  yet  ultimately  fail.  It  can  be  proved  that  no  rational, 
algebraic  formula  for  the  determination  of  prime  numbers  exists,  yet 
many  formulas  have  been  discovered  which  give  prime  numbers  for 
many  consecutive  values  of  n.  Those  who  discovered  such  formulas 
often  believed  them  true  for  all  integral  values  of  n.  For  example,  the 
Chinese,  as  early  as  the  time  of  Confucius,  believed  that  n  is  prime 
if  2«  - 1  —  1  was  exactly  divisible  by  n.  The  least  value  of  n  for  which 
this  theorem  is  not  true  is  341.  Legendre  (1752-1838)  gave  2  n^  +  29 
as  a  formula  for  primes,  and  Euler  (1707~1783)  gave  n^  ■\-  n  +  41. 
Both  of  these  fail  before  n  reaches  50.  Of  course  each  of  these  men 
knew  that  his  formula  is  not  true  generally.  If  the  student  will 
determine  the  least  value  of  n  for  which  either  formula  fails,  he  will 
realize  that  a  large  number  of  numerical  verifications  of  a  theorem 
is  very  far  indeed  from  a  general  proof  of  it.  The  necessity  of  fol- 
lowing several  numerical  verifications  by  mathematical  induction  to 
establish  the  general  truth  of  a  formula  should  then  be  apparent. 


EXAMPLES 

1.  It  has  been  proved  by  the  factor  theorem  that  for  all 
integral  values  of  n,  a"  —  Z*"  is  exactly  divisible  by  a  —  b. 
Nevertheless,  a  proof  by  induction  will  be  given,  since  it 
appears  to  be  slightly  different  from  the  example  on  pages 
26-27,  though  it  is  essentially  the  same. 


252  SECOND  COURSE  IN  ALGEBRA 

Proof.  From  the  work  in  factoring  we  know  that  a"  —  />"  is 
divisible  hy  a  —  b  for  integral  values  of  n  up  to  7  at  least. 

Therefore  a*  —  ^*  is  exactly  divisible  hy  a  —  b  for  all  values 
of  k  from  1  to  7. 

Hence  for  these  values  of  k,  a  —  b  will  exactly  divide 

a  (a*  -  ^,^)  +  ^,*  (a  -  ^>).  .        (1) 

But  (1)  becomes       a* + ^  -  «/>*  +  ab^  -b^^\  (2) 

or  a*  +  i-^»*  +  ^  (3) 

Therefore  (3)  is  exactly  divisible  by  a  —  ^. 

Hence  since  a'  —  b"^  is  exactly  divisible  by  a  —  b,  a^  —  b^ 
is  also.  But  if  a^  —  b^  is  exactly  divisible  by  a  —  b,  so  is 
a*  —  h^,  etc.  Therefore  a"  —  Z*"  is  exactly  divisible  hj  a  —  b 
for  all  positive  integral  values  of  n. 

2.  An  illustration  somewhat  different  from  either  of  those 
given,  and  apparently  very  difficult,  is  the  following: 

Prove  that  (9'*  +  ^  —  8  7i  —  9)  h-  64  gives  an  integral  quotient 
for  all  values  of  n. 

Proof.  If  71  =  1,  2,  3,  and  4,  (9«  +  ^  -  8  ti  -  9)  -^  64  becomes 
respectively  f  |,  ^-^^,  -^|f^,  and  ^^^-^,  the  respective  quotients 
being  1,  11,  102,  and  922. 

We  know,  therefore,  that  (if  c  is  a  properly  chosen  integer) 
for  every  value  of  k  from  I  to  4 

9*+i_87c-9  =  64.c.  (1) 

Then  9(9^  +  i  -  8A^  -  9)=  9- 64.c,  (2) 

and     9(9^  +  i-87c-9)4-64A:  +  64  =  9-64-c  +  647i;  +  64.  (3) 
Whence 

9^  +  2  _  72  ^  _  81  +  64  A;  +  64  =  64(9  c  +  7i;  +  1),  (4) 

9^  +  2  _  8 7^  _  8  _  9  =  64(9  c  +  7c  -f- 1),  (5) 

or  9fc  +  2_8(7c  +  l)-9=64(9c  +  7c  +  l).  (6) 

Since  the  right  member  of  (6)  is  divisible  by  64  the  left 
member  is  also. 


SUPPLEMENTAEY  TOPICS  25B 

But  the  left  member  of  (6)  is  the  same  as  9*+ ^  —  8  A;  —  9,  with 
k  -^1  put  for  k. 

Hence  (9*  +  ^  —  8^  —  9)  ^  64  is  an  integer  when  k  is  re- 
placed by  ^  +  1.  Consequently  (1)  is  true  when  k  =  5.  Then 
it  follows  that  (1)  is  true  when  k  =  6,  and  so  on. 

Therefore  (9"  +  ^  —  8  ti  —  9)  -=-  64  gives  an  integer  for  all 
positive  integral  values  of  n. 

EXERCISES 

Prove  by  mathematical  induction  that 

1.  The  sum  of  the  first  n  integers  is  -(n-\-l). 

2.  The  sum  of  the  first  n  odd  integers  is  r^. 

3.  The   sum   of   the   squares    of   the   first   n   integers   is 
^(»  +  l)(2»  +  l). 

4.  The  sum  of  the  squares  of  the  first  n  odd  integers  is 

^(27i  +  l)(2^-l). 

5.  The  sum  of  the  first  n  integral  powers  of  the  number  2 

is  2  (2^  -  1). 

6.  a  4-  ar  +  cvr'^  -\ \-  ar^-^  =  ^  ^     ~    ^  • 

r  —  1 

7.  3  +  6  +  9  +  ... +  371  =  ^(7^  +  1). 

8.  ic^"  —  ?/2"  is  divisible  by  a?  +  ?/  if  ti  is  a  positive  integer. 

9.  x"  +  T/**  is  divisible  hjx-{-yiinis  odd. 

10.  1  •  2  +  2  •  3  +  3  •  4  +  . . .  +  71  (71  +  1)  =  ^  (71  +  l)(7i  +  2). 

o 

11.  2.5  +  3.6  +  4.7  +  .-.  +  (7^  +  l)(7i  +  4)  =  ^(7i  +  4)(7i  +  5> 

12.  A+A+A+--+    ^ 


1.2'2.3'3.4'  '   n(n-{-l)      n-\-l 

42  +  r  +  102_^-...  +  (3,^  +  l)2  = 
14.  9"  +  ^  —  1  is  exactly  divisible  by  4. 


13.  42  +  7"  +  10^  +  . .  •  +  (3  71  +  ly  =  ^  (6  7^2  +  15  71  +  11). 


254  SECOND  COURSE  IN  ALGEBRA 

15.  Prove  3^"  —  1  is  exactly  divisible  by  8. 

16.  Prove  3^^"  —  8n  —  1  is  exactly  divisible  by  64. 

17.  Prove  9"  +  ^  —  1  is  exactly  divisible  by  8. 

18.  Prove  S  ri^  -\- 15  n  -\- 6  is  exactly  divisible  by' 6. 

19.  Prove  n(n-{-l)(n-^  5)  is  exactly  divisible  by  6. 

20.  Prove  (a  +  hy 

^  I  o  I 

21.  A  pyramid  of  shot  stands  on  a  triangular  base  having  m 
shot  on  a  side.    How  many  shot  are  in  the  pile  ? 

22.  Prom  Exercises  3  and  4  derive  a  formula  for  the  sum 
of  the  squares  of  the  first  n  even  integers. 

139.  Proof  of  remainder  theorem.  This  theorem  is  stated  on 
page  24  as  follows :  If  any  rational  integral  expression  in  x 
be  divided  by  x  —  n,  the  remainder  is  the  same  as  the  original 
expression  with  n  substituted  for  x.  Let  f(x)  denote  any 
rational  integral  function  of  x  as  follows : 

f(x)^ax'' -{-bx^'-'^  +  cx''-^ -\ p.  (1) 

Then  f(k)  =  «A;»  +  hk''-^  +  ck»-^  +  ■ --p.  (2) 

(1)  -  (2),  f(x)  -f(k)  =  ax^  -{-bx«--'+  cx-^  +  •  •  • 

-ak'^-hk^-'^-ck^-^ (3) 

=  a(a;«  -  ^")  +  &(ic"-i  —  A;"-^) 

+  c(ic«-2-A:''-2)H .  (4) 

By  Example  1,  page  251,  each  binomial  in  the  right  mem- 
ber of  (4)  is  divisible  hy  x  —  k.    Denote  the  quotient  by  Q  (x), 
and  the  right  member  of  (4)  may  be  written  (x  —  k)  Q  (x). 
Therefore  f(x)  -  f(k)  =  (x-k)Q  (x).  (5) 

Transposing  f(k)  and  dividing  by  x  —  k,  (5)  becomes 

x  —k         ^  ^      X  —  k  ^  ^ 

That  is,  f{k)  is  the  remainder  when  f(x)  is   divided  by 

{X  -  k). 


SUPPLEMENTARY  TOPICS  255 

140.  Theorems  on  irrational  numbers.  In  order  to  solve  a 
linear  equatioji  or  a  system  of  linear  equations  having  rational 
coefficients,  we  need  use  only  the  operations  of  addition,  subtrac- 
tion, multiplication,  and  division.  When,  however,  we  attempt 
to  solve  the  equation  of  the  second  degree,  x^  =  2,  we  find  that 
there  is  no  rational  number  which  satisfies  it.  This  last  fact 
can  be  proved  if  (as  in  Example  2,  page  252)  we  assume  that : 

An  integral  factor  of  one  member  of  an  identity  between  inte- 
gers is  also  a  factor  of  the  other  member. 

This  is  certainly  true.  For  example,  let  2  a  =  b,  where  a  and 
b  are  integers.  Then  since  2  is  a  factor  of  the  left  member  it 
is  also  a  factor  of  the  right. 

Theorem  1.    No  rational  number  satisfies  the  equation  x^  =  2. 
Proof.   Evidently  no  integer  satisfies  the  equation.    Let  us  make 

the  supposition  that  a  rational  fraction  in  its  lowest  terms,  - ,  satis- 
fies it.    Then  ,  ,  „  ^ 

or  a2  =  2&2.  (2) 

From  (2)  it  is  seen  that  2  is  a  divisor  of  the  left  member  and 
therefore  a  divisor  of  a^,  and  hence  a  divisor  of  a.  Let  us  then  sup- 
pose a  -^  2  =  m,  ov                               _  ^ox 

a  =  2m.  (o) 

(4) 
(5) 
(6) 
Hence  2  must  be  a  divisor  of  h^  and  therefore  of  h.   Then  2  is  a 

divisor  of  both  a  and  h,  which  contradicts  the  hypothesis  that  -  is  a 
rational  fraction  in  its  lowest  terms. 

Therefore  no  rational  number  satisfies  the  equation  x'^  =  2. 

Note.  This  theorem,  when  stated  in  geometrical  language,  asserts 
that  the  hypotenuse  of  an  isosceles  right  triangle  is  not  commensu- 
rate with  the  legs  of  the  triangle.  In  this  form  the  theorem  was 
stated,  and  perhaps  proved,  by  Pythagoras,  about  525  B.C.  The  proof 
given  here  is  found  in  Euclid's  "Geometry,"  and  some  historians 
think  that  it  is  the  very  demonstration  given  by  Pythagoras  himself, 
and  was  inserted  by  Euclid  in  his  book  for  its  historical  interest. 


Then 

«2  =  4  m2 

From  (2)  and  (4), 

4m2  =  262, 

2  ni"  ^  h\ 

256  SECOND  COURSE  IN  ALGEBRA 

Theorem  2.  The  square  root  of  a  rational  number  cannot  he 
the  sum  of  a  rational  number  and  a  quadratic  surd. 

Proof.  Suppose  a:  is  a  rational  number  and  Va  and  V^  are  surds. 
Then,  if  possible,  suppose 

-s/a=-\/b±x.  (1) 
Squaring  each  member  of  (1), 

a  =  b  +  x^±2x  Vb.  (2) 

Solving  (2),  Vi  =  ±  ^-b-^' .  (3) 

Ji  X 

But  (3)  is  impossible,  for  it  asserts  that  a  surd  equals  a  rational 
number. 

Therefore  ■y/a  ^  -y/b  ±  a;  if  -y/a  and  Vi  are  surds. 

Theorem  3.  If  each  tnember  of  an  equation  consists  of  a 
rational  number  and  a  quadratic  surd,  then  the  rational  parts 
are  equal  and  the  irrational  paHs  are  equal. 

Proof  Let  a  Jr  ^  =  c -\-  Vd.  (1) 

If  possible,  suppose  c  =  a  ±  x.  (2) 

Then  a  +  Vb  =^  a  ±  x  +  Vd,  (3) 

or  ■Vb  =  ±x  +  Vd.  (4) 

But  (4)  by  the  preceding  theorem  is  impossible. 

Consequently  a  =  c,  and  hence  from  (1),  -wf)  =  Va. 

Therefore,  if  a  +  \b  =  c  +  -yd,  a  =  c  and  -wb  =  -y/d. 

141.  Cube  root  of  algebraic  expressions.  Since  by  actual  mul- 
tiplication 

(t-\-uy  =  t^-\-St^u-^Stu^-\-u% 

a  careful  inspection  of  the  expression 

t^  -\-Zt^u  +  3tu^  +  1^ 

will  enable  one  to  extract  the  cube  root  of  any  polynomial 
which  is  a  perfect  cube ;  for  the  extraction  of  cube  and  other 
roots  is  not  a  mysterious,  unreasonable  process,  but  merely  an 
intelligent  undoing  of  the  work  of  multiplication.  We  see  that 
the  first  term  of  the  result  is  the  cube  root  of  the  first  term  of 
the  polynomial  ^.  The  second  term  of  the  cube  root,  u,  can  be 
obtained  by  squaring  t,  multiplying  it  by  3,  and  dividing  the 


SUPPLEMENTARY  TOPICS  25T 

result  as  a  trial  divisor  into  the  second  term  of  the  polynomial, 
thus  obtaining  u.  Since  f  +  ^  fu  -{-^tu'  ■]-  u^  =  ^'  +  (3  f  + 
3  tu  4-  u^)  u  =  {t-{-  tif,  we  may  then  form  the  complete  divisor 
as  indicated  by  the  trinomial  in  parenthesis.  A  systematic 
arrangement  of  the  work  follows  : 

Example  1 .  t^  +  ^f^u  +  ^tu^-^-u^ \t -\- u 


Trial  divisor,  S -1^=31^ 

Second  term  of  root, 
St^u^3t'^  =  U 
Complete  divisor,  St^  +  3tu  +  u^ 


St^u-\-Stu^-{-u^ 


tH  +  3tu^^u^={3t'^+3tu-\-  u^)u 


54a2x  +  36ax2-8xa 


Example  2.    Extract  the  cube  root  of  27  a^  —  8  x^  +  36  ax^  —  54  a%. 
Solution:  27 g^  -  54 a^x  +  36 ax^  -  8 x^ |3 a  -  2 x 

t3  =  (3  a)3  =  27  g^ 

Trial  divisor,  3 i^  :^  3  •  (3 g)^  =  27 g^ 
Second  term  of  root,  u,  equals 
-  54  g2x  ^  27  g2  =  -  2  X 

3  ^u  =  3  •  3  g  (- 2  x)  =  -  18  gx 
u'^=  (2x)2  =  4x2 

Complete  divisor,  27  a^  —  18  gx  +  4  x^ 
-2x 


54  g%  +  36  gx2 


54  g2x  +  36  gx2  -  8  x3 


The  student  should  note  particularly  the  form  of  the  trial 
divisor  and  of  the  complete  divisor.  They  are  very  important 
in  extracting  the  cube  root  of  a  polynomial  or  of  an  arithmet- 
ical number. 

If  t  in  the  preceding  example  be  replaced  by  the  binomial 
h  -\-  t,  we  obtain 

[(h  +  t)-{-  uf  =  (h  +  ty  +  S  (h  +  tyic  +  3  (7^  +  t)  u''  +  11^ 

If  this  last  were  expanded  fully,  we  would  obtain  a  polyno- 
mial of  ten  terms  which  would  be  a  perfect  cube.  Its  cube 
root  could  be  obtained  as  before,  first  obtaining  h  and  then  t. 
Then  we  could  regard  A  -t-  ^  as  a  single  term,  form  the  trial 
divisor  and  the  complete  divisor  as  before,  and  obtain  the  third 
term  of  the  root.  Thus  the  method  may  be  extended  to  any 
polynomial  of  more  than  four  terms  which  is  a  perfect  cube. 


258  SECOND  COURSE  IX  ALGEBRA 

The  method  just  illustrated  may  be  stated  in  the 

Rule.  Arrange  the  terms  of  the  polynomial  according  to  the 
powers  of  some  letter  in  it. 

Extract  the  cube  root  of  the  first  term.  Write  the  result  as 
the  first  term  of  the  root,  and  subtract  its  cube  from  the  given 
polynomial. 

Square  the  part  of  the  root  already  found  and  multijdy  the 
result  by  3  for  a  trial  divisor.  Divide  the  first  tervi  of  this 
product  into  the  first  term  of  the  remainder,  and  write  the  quo- 
tient as  the  second  term  of  the  root. 

Annex  to  the  trial  divisor  three  times  the  product  of  the  first 
term  and  the  second  term  of  the  root,  and  the  square  of  the 
second  term  also,  thus  forming  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  second  terTn  of  the  root, 
and .  subtract  the  result  from,  the  remainder. 

If  terms  of  the  polynomial  still  remain,  square  the  part  of 
the  root  already  found,  and  multiply  the  result  by  3  for  a  trial 
divisor.  Divide  the  first  term  of  the  trial  divisor  into  the  first 
term  of  the  remainder,  and  write  the  quotient  as  the  third  term 
of  the  root,  form  the  complete  divisor,  and  proceed  as  before  until 
the  process  ends,  or  until  the  required  number  of  terms  have 
been  obtained. 

EXERCISES 

Extract  the  cube  root  of : 

1.  x^  +  ^x'  +  dx^l.  2.  ^x^-12x^  +  ^x-l. 

3.  27  ic»  +  27  xhj  +  9  £c/  +  y\ 

4.  64  a«  -  144  a'c  +  108  ac''  -  27  c\ 

5.  cci2  -  15  rr^o  -f  75  ic«  -  125  x\ 


e_3^      3^_1  ^_^_£!_l1 

®-  "^  2    "^    4         8'  c«       c^      a''      a^' 

^'  27      2>^  x'      x''  ^-  ^    "^     «2  a        a'' 

10.  «» -f-  Z»8  -  1  -  3  a^  _  3  ^2  _  6  ^^  _^  3  ^2^  _^  3  ^^^2  _^  3  ^  _l_  3  ^ 


SUPPLEMENTARY  TOPICS 


259 


11.  x^-^S-  9x^+  66a^2  _  3g^  _|_  33^4  _  ^3^3 

12.  Find  the  sixth  root  of  x^  -12x'  -\-64.-  192  x  +  24.0  x^ 
-^60x^-160x'. 

13.  Find  the  first  three  terms  in  the  cube  root  of  1  +  3  ic. 

142.  Cube  root  of  arithmetical  numbers.  The  process  of  ex- 
tracting tlie  cube  root  of  an  arithmetical  number  does  not 
differ  greatly  from  the  method  of  extracting  the  cube  root 
of  any  polynomial.  The  formula  for  the  complete  divisor, 
3f  +  Stu-\-  u^,  can  be  used  to  guide  the  important  steps  in 
the  work.  The  first  step,  however,  is  pointing  off,  the  reason 
for  which  appears  from  a  study  of  the  following  table : 


n  = 

1 

10 

100 

1000 

rfi  = 

1 

1000 

1,000,000 

1,000,000,000 

From  this  table  it  is  obvious  that  the  cube  root  of  an  integral 
number  of  three  digits  or  less  must  contain  only  one  digit  on 
the  left  of  the  decimal  point.  Similarly,  we  see  that  the  cube 
root  of  an  integral  number  containing  four,  five,  or  six  digits 
contains  two  digits  on  the  left  of  the  decimal  point ;  and  the 
cube  root  of  an  integral  number  of  seven,  eight,  or  nine  digits 
contains  three  digits  on  the  left  of  the  decimal  point.  Hence 
in  cube  root  we  find  it  convenient  to  begin  at  the  decimal  point 
and  point  off  the  number  in  periods  of  three  figures  each,  —  to 
the  left  if  the  number  is  integral,  to  the  right  if  it  is  decimal ; 
to  both  the  left  and  right  if  the  number  is  part  integral  and 
part  decimal.  There  may,  of  course,  be  an  incomplete  period 
on  the  left.  Zeros  should  be  used  to  complete  any  partial 
period  on  the  right. 

If  we  now  imagine  (t  +  uY  to  be  a  number  consisting  of  a 
tens  and  a  units  digit,  we  may  translate  ^^  +  3  fit  -+-  3  tu^  -f  u^ 
thus  :  the  cube  of  the  tens  +  3  times  the  square  of  the  tens  times 
the  units  +  S  times  the  tens  times  the  square  of  the  units  -f-  the 
cube  of  the  units. 


260 


SECOND  COURSE  IN  ALGEBRA 


Now  subtracting  ^  from  the  polynomial,  we  may  write  the 
other  three  terms  thus  :  (3  ^^  +  3  tu  -\-  u^)  u.  Here  the  trinomial 
in  parenthesis  is  the  complete  divisor.  The  process  of  ex- 
tracting the  cube  root  of  50,653  follows  : 


f3  ^         (30)8  ^ 

Trial  divisor,  3  «2  =  3  •  (30)2  =  2  700 

Second  term  of  root,  «,  23653  -^  2700  =  7  + 

3«M  =  3.30.7=    630 


60^653  |30+  7 
27  000 


37 


72        = 


49 


23  653 


23  653  =  3  379  X  7 


Complete  divisor,       3  «2  +  3  <u  +  u2  =  3  379 

To  obtain  the  second  term  of  the  root,  we  divided  23,653  by 
2700,  which  gave  almost  exactly  the  number  8.  But  since  the 
trial  divisor  2700  must  be  increased  by  Ztu  and  i^  to  form 
the  complete  divisor,  a  moment's  thought  showed  that  8  was  too 
great.  This  means  that  the  trial  divisor  is  really  a  trial  divisor, 
and  its  use  does  not  give  us  with  certainty  the  next  term  of 
the  root.  A  little  experience  will  enable  one  to  look  ahead  and 
decide  mentally  on  the  next  root  figure.  If  one  decides  on  a 
root  figure  either  too  great  or  too  small,  the  product  of  the 
complete  divisor  and  this  root  digit  will  be  too  gre'at  or  too 
small  and  the  subsequent  work  will  show  the  error. 

Since  374,  for  example,  may  be  regarded  as  37  tens  plus  4 
units,  the  process  just  illustrated  may  be  applied  to  a  number 
whose  cube  root  contains  three  digits,  as  follows : 


t3  =           (800)3  = 

644'972'544|800  +  60  +  4  =  864 
512  000  000 

3^2=       3(800)2  =  1920000 
Second  term  of  root,  u,  equals 
132  972  534  -  1  920  000  =  60  + 

3<M  =  3(800)(60)=     144  000 
m2  =            (60)2  =         3  600 

132  972  644 

3i2  +  3tM  +  w2                    =2  067  600 

124  056  000  =  2  067  600  x  60 

3<2=    3(860)2  =  2  218  8( 
Third  term  of  root, 
8  916  534  -4-  2  218  800  =  4  + 

3«M  =  3(860)4=       10  3. 
m2  =             42  = 

K) 

20 
16 

8  916  544 

3 12  +  3  «u  +  u2  =  2  229  1, 

36 

8  916  544  =  2  229  136  x  4 

INDEX 


Abel,  247 

Addition,   algebraic,   2 ;    of   frac- 
tions, 35 
Ahmes,  166 
Alternation,  208 
Antecedent,  207 
Antilogaritlim,  193 
Argand,  231 
Axiom,  42 
Axioms,  42 

Base,  189 

Bernouilli,  John,  180 
Binomial   Theorem,   243 ;    extrac- 
tion of  roots  by,  245 

Characteristic,  189 

Coefficients,  detached,  8 

Complex  number,  223;    graphical 

representation  of,  230 
Confucius,  251 
Conjugate  radicals,  100 
Conjugates,  226 
Consequent,  207 
Constant,  212 
Coordinates,  55 

Determinant,  62 

Determinants,  solution  by,  63 ;  of 

third  order,  71 
Diophantos  of  Alexandria,  67 
Discriminant,  234 
Division,  rule  for,  5  ;  synthetic,  10; 

logarithmic,  187  ;  by  logarithms, 

196 

Elimination,  57 

Ellipse,  138 

Equations,  exponential,  203  ;  for- 
mation of,  with  giyen  roots,  238  ; 
homogeneous,  144 ;  indetermi- 
nate, 65  ;  irrational,  128  ;  kinds 
of,  42  ;  solution  of,  by  factoring, 
31 ;  with  imaginary  roots,  227 


Euler,  247,  251 

Evolution,  by  logarithms,  198  ;  log- 
arithmic, 188 

Exponents,  fractional,  90 ;  funda- 
mental laws  of,  89;  meaning 
of,  6 

Extremes,  207 

Factor,  highest  common,  31 
Factor  Theorem,  25 
Factorial  notation,  243 
Factoring, '  general  directions  for, 

28 
Fractions,  addition  and  subtraction 

of,  35  ;  complex,  38  ;  equivalent, 

34 
Function,  110 ;  graph  of,  110 ;  graph 

of  a  cubic,  110 

Gauss,  116,  223,  231 

Girard,  239 

Graph,  of  cubic  function,  110 ; 
of  function,  110;  of  quadratic 
equation  in  two  variables, 
135 

Graphical  explanation  of  loga- 
rithms, 183 

Graphical  interpretation  of  pure 
imaginaries,  229 

Graphical  method  of  extracting 
roots,  84 

Graphical  representation  of  a  com- 
plex number,  230 

Graphical  solution  of  a  linear  sys- 
tem, 54 

Graphical  solution  of  an  equation 
in  one  unknown,  113 

Graphical  solution  of  a  quadratic 
system  in  two  variables,  139 

Graphical  solution  of  two  linear 
equations,  55 


Harriot,  239 
Hyperbola,  136 


263 


264 


SECOND  COURSE  IN  ALGERRA 


Imaginaries,  addition  and  subtrac- 
tion of,  223 ;  division  of,  226 ; 
equations  with  imaginary  roots, 
227 ;  factors  involving,  228 ; 
graphical  interpretation  of  pure, 
229  ;  note  on  the  use  of,  232 

Imaginary  roots,  112 

Index,  93 

Induction,  mathematical,  249 

Infinity,  176 ;  definition  of  the 
term  "infinite,"  176 

Interpolation,  192 

Inversi<m,  208 

Involution,  by  logarithms,  197 ; 
logarithmic,  188 

Irrational  numbers,  theorems  on, 
255 

Klein,  Professor  Felix,  116 

Legendre,  251 

Leibnitz,  Gottfried  Wilhelm,  64 

L' Hospital,  180 

Limits,  176;  definition  of,  176 

Logarithms,    antilogarithm,    193 ; 

graphical   explanation  of,   183 ; 

interpolation,  192 ;  table  of,  200 

Mantissa,  189 

Mean  proportional,  207 

Means,  207 ;  arithmetical,  161 ;  geo- 
metrical, 168 

Merehiston,  Lord  of  (John  Napier), 
190,  206,  246 

Multiple,  lowest  common,  32 

Multiplication,  by  logarithms,  195  ; 
law  of,  6  ;  logarithmic,  187 

Napier,  John  (Lord  of  Merehiston), 
190,  206,  246 

Kewton,  247 

Origin,  55 

Parabola,  135 

Progressions,  arithmetical,  159 ; 
geometrical,  166 


Proportion,  207 

Proportional,  fourth,  208  ;  mean, 
207 ;  third,  208 

Quadratic  equation,  character  of 
the  roots  of ,234  ;  number  of  roots 
of,  238 ;  relation  between  the 
roots  and  the  coefficients  of, 
236 

Quadratics,  factors  of,  20,  240 

Quantity,  211 

Radicals,  93 ;  addition  and  sub- 
traction of,  97  ;  division  of,  100  ; 
nmltiplication  of,  97 ;  simplifica- 
tion of,  95 

Radicand,  93 

Ratio,  166,  207 

Rational  number,  93 

Remainder  Theorem,  24  ;  proof  of, 
254 

Root,  cube,  of  algebraic  expressions, 
256 ;  of  arithmetical  numbers, 
259 

Root,  square,  of  arithmetical  num- 
bers, 86  ;  of  polynomials,  83  ;  of 
surd  expressions,  102 

Roots,  imaginary,  112 

Series,  geometrical,  169 ;  infinite 
geometrical,  171 ;  sum  of,  163 

"Slide  rule,"  199 

Subtraction,  of  fractions,  35 ;  of 
polynomials,  3 

Surd,  93 

Systems,  determinate,  68  ;  equiva- 
lent, 147  ;  general  linear,  in  three 
variables,  72 

Table  of  (Cubes  and  squares,  262 

VariaHe,  212 

Variation,   211  ;   direct,   212 ;    in- 
verse, 213;  joint,  215 
Vieta,  239 

Wessel,  231 


14  DAY  USE 

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LD  21A-507n-4,'59 
(A1724slO)476B 


General  Library 

University  of  California 

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